Questions on Algebra: Quadratic Equation answered by real tutors!

Tutors Answer Your Questions about Quadratic Equations (FREE) Get help from our free tutors ===> Algebra.Com stats: 1684 tutors, 337023 problems solved View all solved problems on Quadratic_Equations -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. If you need immediate math help from PAID TUTORS right now, click here. (paid link) Click here to see problems with only links to answers, all on one page Question 570883: Ticket Sales Living in or by a metropolitan area has certain advantages. Entertainment opportunities are almost boundless in a major city. Events occur almost every night, from sporting events to the ballet. Tickets to these events are not available long; and quantity of tickets demanded can often be modeled by quadratic equations. Application Practice Answer the following questions. You must use Equation Editor or MathType when writing mathematical expressions or equations. Working in a new MS Word file, provide solutions and answers to all problems, clearly labeling your work. You must show your steps or provide verbal explanations (where appropriate) to receive full credit. Use textbook examples as your guide as to what level of detail is expected. 1. Suppose you are an event coordinator for a large performance theater. One of the hottest new Broadway musicals has started to tour and your city is the first stop on the tour. You need to supply information about projected ticket sales to the box office manager. The box office manager uses this information to anticipate staffing needs until the tickets sell out. You provide the manager with a quadratic equation that models the expected number of ticket sales for each day n. ( is the day tickets go on sale, Day 1). [Note: Function f(n) “maps” the days (dates) when tickets are sold to the corresponding number of tickets sold on each specific day (date), so f(15) would denote the number of tickets sold on the 15th day of ticket sales] a. Does the graph of this equation open up or down? How did you determine this? b. Describe what happens to the tickets sales as time passes. c. Use the quadratic equation to determine the last day that tickets will be sold. Note. Write your answer in terms of the number of days after ticket sales begin. d. Will tickets peak or be at a low during the middle of the sale? How do you know? After how many days will the peak or low occur? e. How many tickets would be sold on Day 4? On Day 13? On Day 32? f. How many tickets will be sold on the day when the peak or low occurs? g. What is the point of the vertex? How does this number relate to your answers in parts d. and f? h. How many solutions are there to the equation ? How do you know? i. What do the solutions represent? Is there a solution that does not make sense? If so, in what ways does the solution not make sense? j. (Optional – advanced) How many tickets in total will be sold during the entire period when tickets are sold? [Hint: One can, of course, take this problem “heads on”, calculating the number of tickets sold on each day that tickets are sold (e.g. for all n when f(n)>0). However, this will involve way too much work, as you probably have seen in e. We need to “speed up” the process. How? Well, if a constant number of tickets (e.g. 100 tickets) were sold on days 1 through k, we know that the total number of tickets sold would be 100*k. You should use your answer to c. above as your actual k. What if ticket sales were proportional to the day number, e.g. 1 ticket sold on Day 1 and k tickets sold on Day k? Formula for the sum of an arithmetic progression (specific to this case) would yield the number of tickets to be . Now if the proportionality coefficient were to be not 1 but some “b”, then the formula would simply be: (do you see why?) The most challenging is probably the formula for the total number of tickets sold if ticket sales were directly proportional to the square of the day number: . Try this formula to see if it works for the first few sums of squares (e.g. 1+4+9, here k would be 3). Of course, if there was a proportionality coefficient different from 1, say “a”, then the formula would simply be: Now, all you have to do would be to understand that the quadratic function’s three separate elements may be evaluated separately using the above formulas for the entire domain of days when tickets are sold, and the total can be arrived at much quicker than taking the problem “head on”] I am totally confused on this problem and not matter what I did I cannot figure it out. Please help. Answer by richard1234(4789)   (Show Source): You can put this solution on YOUR website! Try to limit your post to one or two questions; here, you posted two questions that have 10 or 12 parts and most tutors won't solve all of them for you. Plus, you didn't provide a quadratic for #1. For #2, you can approximate it by adding up each f(n), 0 <= n <= d, where f(n) is the function and d is the last day tickets are sold. However, unless you have a graphing calculator that can do summations, this will be quite a task. Another way to approximate it is to take the integral of f(n): Note the "approximately equal," as the integral assumes f(n) is continuous instead of day-to-day. Only use this method if you know calculus. Question 571016: my homework question reads: an object is thrown upward from the top of a 256ft tower at a velocity of 96ft/sec. Find the height after 3 seconds. I know I need to use the quadratic equation, but I don't know what to do next. 3t=-16t^2+96t+256 16t^2-93t-256=0 Thanks for your help! Answer by KMST(576)   (Show Source): You can put this solution on YOUR website! According to Physics, the height as a function of time in seconds should be After 3 seconds, the object's height is The quadratic function has a maximum and its graph (h versus t) is a parabola. To figure out when it reaches its highest altitude, find the axis of symmetry of the parabola: --> So the object reaches its maximum height at 9 seconds. That height is To find out when it hits the ground (), you solve the quadratic equation or the equivalent Question 570939: x^2-6x-28=0 Answer by jim_thompson5910(21667)   (Show Source): You can put this solution on YOUR website! Start with the given equation. Notice that the quadratic is in the form of where , , and Let's use the quadratic formula to solve for "x": Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . Simplify the square root (note: If you need help with simplifying square roots, check out this solver) Break up the fraction. Reduce. or Break up the expression. So the solutions are or Question 570835: I am having a hard time solving this equation quadratic in form: 9x(to the 4th power)=85x(squared)-196. I know it then becomes 9x(to the 4th power)-85x(squared)+196=0. Then u = x squared. So then it becomes 9u squared-85u+196=0 Now I'm stuck on how to solve the equation from here because I don't think I factor here??? Answer by mananth(10539)   (Show Source): You can put this solution on YOUR website! u=x^2 In such problems you can solve by quadratic formula or completing the squares Take the square root on both sides 3u-85/6 = -/-(13/6) 3u=85/6 + 13/6 3u=98/6 u=98/18 OR 3u= 85/6 -13/6 3u=72/6 u=72/18 u=4 when u = 4 , x=+/-2 when u=98/18 =====> 49/9===>x=sqrt(49/9)====>+/- 7/3 m.ananth@hotmail.ca Question 570705: I need to solve : f(x)=(1)/(3) (x+4)^(2)+3 for the vertex, and the line of symmetry..Have been working on this one problem for about 20 minutes now. Answer by scott8148(5879)   (Show Source): You can put this solution on YOUR website! the general vertex equation is ___ f(x) = a(x - h)^2 + k ; where (h,k) is the vertex so the axis of symmetry would be ___ x = h Question 570645: Solve the equation by factoring. (2x/x+3)+(5/x)-4=(18/x^2+3x) -------------------------------- I have gotten this far with the problem.... LCM = X(x+3) thus I have 2x^2+5x+15-4x^2-12=18--->-2x^2+5x+3=18--->-2x^2+5x-15=0 I think my problem is the (-2x^2) which I got from 2x^2-4x^2 and the (-4x^2) is from multiplying the LCM (x(x+3))to -4. Could you show me where my error is? Thank you Kenneth Answer by Earlsdon(6098)   (Show Source): You can put this solution on YOUR website! Solve: Simplify. The LCD is so we have: Subtract 18 from both sides. Multiply through by -1 to get: Factor. Apply the zero product rule: or so... or Notice: The left side becomes... which, when simplified becomes: = Question 570512: please help me solve this problem 2x to the second power -3=1 Answer by jillfired(1)   (Show Source): You can put this solution on YOUR website! Since -3 does not contain the variable variable to solve for, move it to the right-hand side of the equation equation 2x^2=3=1 A mathematical statement that says two expressions have the same value; any number sentence with an = . by adding 3 to both sides. Add 1 to 3 to get 4. 2x^2=4 Divide each term in the equation by 2 2x^2/2 = 4/2 Simplify the left side by cancelling out common factors: A mathematical statement that says two expressions have the same value; any number sentence with an = . by 2. x^2=4/2 simplify the right hand side x^2=2 eliminate exponent by taking sqrt of both sides so x= sqrt+/-2 Question 570434: the value of d in this equation: d-34.8=5.3 Answer by solver91311(12117)   (Show Source): You can put this solution on YOUR website! Add 34.8 to both sides. John My calculator said it, I believe it, that settles it Question 569602: 1/a+1/b+1/x=1/a+b+x ,a#0,b#0,x#0,x#-(a+b). Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! 1/a+1/b+1/x=1/a+b+x ,a#0,b#0,x#0,x#-(a+b). --------- What does a#0 mean? What do you want to do? Question 569574: Find Y in the following equation: Answer by JBarnum(1826)   (Show Source): You can put this solution on YOUR website! Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=56 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 3.87082869338697, 0.129171306613029. Here's your graph: Question 569421: Q. Obtain the values of k such that x^2-6x+k=0 has real roots. is it to solve by making b^2-4ac=0 i got the answer as k=9 also Calculate the value of if the difference between the roots of x^2-6x+k=0 is 3 i got the answer as k=27/4 please can u help me to varify my answers.. i will be happy to forward u the steps which i prepared in word. to ur email. Thanks Answer by richard1234(4789)   (Show Source): You can put this solution on YOUR website! b^2 - 4ac only has to be greater than or equal to 0, not necessarily equal to 0, so check the first problem. For the second one, the roots are For the roots to differ by 3, you'll need the sqrt(36-4k) to equal 3. This is because +3/2, -3/2 differ by 3 (you'll get something like 9/2, 3/2). So we have , k = 27/4, correct. Question 569254: How do you know if a quadratic equation will have one, two, or no sloution? Answer by richard1234(4789)   (Show Source): You can put this solution on YOUR website! All quadratics will have two (complex) solutions. However, to determine if they're real or if there are duplicate roots, you'll have to take the discriminant, defined as delta = b^2 - 4ac. Recall the quadratic formula If b^2 - 4ac > 0, then the square root will be defined and you will have two real solutions. If b^2 - 4ac = 0, then the square root will be equal to 0, and you'll have a duplicate root (-b/2a). If b^2 - 4ac < 0, then the square root will be a complex number and you'll have two complex solutions (by complex, I mean Im(x) is not equal to 0). Question 569266: How do you find a quadratic equation if you are only given the solution? Is it possible to have different equadratic equations with the same solution? Answer by richard1234(4789)   (Show Source): You can put this solution on YOUR website! You can find the quadratic equation given the roots. Suppose the roots are p and q (quadratics will only have two roots). Then x-p and x-q are factors of the quadratic, and the quadratic will look like There are infinitely many quadratics that satisfy, but they will all differ by a constant factor, i.e. where C is any constant not equal to 0. Question 569428: Write a quadratic equation in the variable x having the given numbers as solutions. Type the equation in standard form, ax2 + bx + c = 0. Solutions: -9 and 6. Answer by stanbon(48510)   (Show Source): You can put this solution on YOUR website! Write a quadratic equation in the variable x having the given numbers as solutions. Type the equation in standard form, ax2 + bx + c = 0. Solutions: -9 and 6. ----- y = (x+9)(x-6) ---- y = x^2 + 3x - 54 ================================== Cheers, Stan H. Question 569221: y=x-1 y=-x=1 Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! y=x-1 y=-x=1 Too many equal signs. ----- Question 569208: Which of the binomials below is a factor of this expression? 25x2 - 9y2 z2 Answer by stanbon(48510)   (Show Source): You can put this solution on YOUR website! Which of the binomials below is a factor of this expression? 25x^2 - 9y^2*z^2 -------------------- = (5x-3yz)(5x+3yz) ======================= Cheers, Stan H. Question 569093: how do i find the solution of x2-x+3/16=0 by factoring? Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! how do i find the solution of x2-x+3/16=0 by factoring? -------------- (x - 1/4)*(x - 3/4) = 0 Question 569115: How do you solve this quadratic equation explain process. te{5/3,3/2} Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! How do you solve this quadratic equation explain process. te{5/3,3/2} ----------------- What equation? Question 569015: One leg of a right triangle is 7 miles longer than the other. The hypotenuse is equal to 13 miles. How long is the shorter leg? The correct answer is 5 miles. please help me see how to do this! Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! One leg of a right triangle is 7 miles longer than the other. The hypotenuse is equal to 13 miles. How long is the shorter leg? ---------------- Question 569018: One leg of a right traingle is 3 inches shorter than the hypotenuse. The other leg is equal to 8 inches. How long is the first leg? The correct answer is 55/6 inches, but I can't figure out how ot get it. Please help me! Answer by stanbon(48510)   (Show Source): You can put this solution on YOUR website! One leg of a right traingle is 3 inches shorter than the hypotenuse. The other leg is equal to 8 inches. How long is the first leg? leg 1 : 8 hypotenuse : x leg 2 : x-3 ---- Pythagoras: 8^2 + (x-3)^2 = x^2 64 + x^2-6x+9 = x^2 6x = 73 x = 73/6 (hypotenuse) --- x-3 = (73/6)-(18/6) = 55/6 =========== Cheers, Stan H. Question 568822: whats the quadratic equation for the table x=1 and 3 when y =0 for both of them? Answer by stanbon(48510)   (Show Source): You can put this solution on YOUR website! whats the quadratic equation for the table x=1 and 3 when y =0 for both of them? --- Those x-value are the x-intercepts or roots. --- Eqution: y = (x-1)(x-3) --- y = x^2 - 4x +3 ================= Cheers, Stan H. ================= Question 568702: Model with a quadratic equation, then solve. Find the dimensions of a square picture that make the area of the picture equal to 75% of the total area enclosed by the frame. The total length of the side is 12inches, including the frame. I thought y=axsquared for the equation 144=0.75x2 192=x2 x=13.9 But this is larger than the length of the picture frame??? Answer by KMST(576)   (Show Source): You can put this solution on YOUR website! You've got it backwards. If the length of the side of the picture frame is inches, the total area of frame plus picture is square inches = square inches. If inches is the length of the side of the square picture, the area of the square picture is square inches. Since that is 75% of the square inches of total area, --> Solving: --> --> --> --> --> --> --> Question 568741: x^2+3x+8=5 Answer by jim_thompson5910(21667)   (Show Source): You can put this solution on YOUR website! Start with the given equation. Subtract 5 from both sides. Combine like terms. Notice that the quadratic is in the form of where , , and Let's use the quadratic formula to solve for "x": Start with the quadratic formula Plug in , , and Square to get . Multiply to get Subtract from to get Multiply and to get . Simplify the square root (note: If you need help with simplifying square roots, check out this solver) or Break up the expression. So the solutions are or ------------------------------------------------------------------------------------------------- If you need more help, email me at jim_thompson5910@hotmail.com Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you Jim ------------------------------------------------------------------------------------------------- Question 568662: 3x^2-x-5=0 Answer by jim_thompson5910(21667)   (Show Source): You can put this solution on YOUR website! Start with the given equation. Notice that the quadratic is in the form of where , , and Let's use the quadratic formula to solve for "x": Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . or Break up the expression. So the solutions are or ------------------------------------------------------------------------------------------------- If you need more help, email me at jim_thompson5910@hotmail.com Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you Jim ------------------------------------------------------------------------------------------------- Question 568627: I have no idea why I got a certain answer. This is my problem: 5/6x+30=x then I multiplied both by 6 and added them together to get 185. Am I correct? If so, how does that answer work? Answer by nyc_function(2626)   (Show Source): You can put this solution on YOUR website! Nice try but you needed to multiply all terms by 6x. If your question is (5/6x) + 30 = x, then your answer will be positive and negative for x. (5/6x) + 30 = x 6x * (5/6x) + 30 * 6x = x * 6x 5 + 1800x = 6x^2 6x^2 - 1800x -5 = 0 This is quadratic equation. Using the quadratic formula (I assume that you know how to use this formula), I got the following two answers for x. x = [90 + sqrt{8130}]/6 and x = [90 - sqrt{8130}]/6 NOTEL sqrt = square root for short If you need to learn how to use the quadratic formula, watch this video clip. http://www.youtube.com/watch?v=UgIF7Q7lAuc Question 568520: how do you completely factor y5-26y Answer by jim_thompson5910(21667)   (Show Source): You can put this solution on YOUR website! Question 568196: a pool is to be 30 feet wide and 90 feet long. A walkway is to be built around the entire pool. if the combined area cannot exceed 4000 square feet what are the possible widths of the walkway? Answer by mananth(10539)   (Show Source): You can put this solution on YOUR website! Area of pool = 30* 90 = 2700 ft^2 ( rectangle) Area of pool + walkway = 4000 ft^2 area of walkway = 1300 ft^2 let the width of walk way be x (30+2x)(90+2x)=4000 2700+60x+180x+4x^2=4000 4x^2+240x+2700=4000 /4 x^2+60x+675=1000 x^2+60x-325=0 x^2+65x-5x-325=0 x(x+65)-5(x+650=0 (x+65)(x-5)=0 x= 5 m The width has to be less than 5 m Question 566230: how do you solve the equation 4p^2-90=-9p? Answer by jim_thompson5910(21667)   (Show Source): You can put this solution on YOUR website! Start with the given equation. Get every term to the left side. Rearrange the terms. Notice that the quadratic is in the form of where , , and Let's use the quadratic formula to solve for "p": Start with the quadratic formula Plug in , , and Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the solutions are or Question 567835: How do you get the quadratic equation if you are given a table of values? Answer by issacodegard(60)   (Show Source): You can put this solution on YOUR website! You must be given at least 3 points, call them (x1,y1),(x2,y2),(x3,y3). Then we know the equation has the form f(x)=ax^2+bx+c. We get a system of 3 equations in 3 unknowns. Solving for a,b,c in, gives you the equation for f(x). I'll give an example, If we know that three points are (-1,-2),(1,4),(2,13). Then we set up three equations about a,b,c that we know: So we have, We need to solve for a,b,c. So, we subtract eqn1 from eqn2 to get, Divide eqn2 by 2 to get b=3. Then, we have So, Subtract the first eqn from the second to get, Divide the second eqn by 3 to get a=2. Then we have So, Therefore we find that Question 567780: The quadratic equations are difficult. I need help with this equation. Write a quadratic equation having the given numbers as solutions - sqrt5 and 8sqrt5. Answer by Alan3354(21550)   (Show Source): You can put this solution on YOUR website! Write a quadratic equation having the given numbers as solutions - sqrt5 and 8sqrt5. ----------- Question 567571: Tell whether y=-3x^2+18x-20 has a minimum value or a maximum value?Then find the minimum or maximum value. Answer by lwsshak3(2907)   (Show Source): You can put this solution on YOUR website! Tell whether y=-3x^2+18x-20 has a minimum value or a maximum value?Then find the minimum or maximum value. ** Second degree equations are parabolas. Standard form of equation for a parabola which opens downwards: y=-A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the slope of the curve. The negative lead coefficient means the parabola opens downward and therefore, has a maximum value. y=-3x^2+18x-20 complete the square y=-3(x^2-6x+9)-20+27 y=-3(x-3)^2+7 .. given parabola has a maximum=7 at x=3 Question 567351: A box with an open top is constructed from a rectangular piece of cardboard which is initially 30 inches long and 23 inches wide. The box will be formed by removing squares of length x inches from each corner and then folding up the sides (as shown below). Express the volume (V) of the box as a function of x. I've been trying to figure out this problem for a while, any help is greatly appreciated!!! Thanks! Answer by ankor@dixie-net.com(12684)   (Show Source): You can put this solution on YOUR website! A box with an open top is constructed from a rectangular piece of cardboard which is initially 30 inches long and 23 inches wide. The box will be formed by removing squares of length x inches from each corner and then folding up the sides (as shown below). Express the volume (V) of the box as a function of x. : From the information, the box dimensions will be: (30-2x) by (23-2x) by x (Height) : therefore the volume: V(x) = x*(30-2x)*(23-2x) FOIL V(x) = x(690 - 60x - 46x + 4x^2) ; which is usually written V(x) = x(4x^2 - 106x + 690) ; multiply by the height V(x) = 4x^3 - 106x^2 + 690x Question 567353: A flag is raised while an onlooker watches from a distance of 21 feet away from the base of the flag pole (see the figure below). The flag rises vertically at a rate of 8 inches per second. Let t denote the time (in seconds) after the flag begins to rise (For simplicity, assume that when the flag begins to rise it is 0 inches above the ground). Express the distance d (in feet) between the flag and the onlooker as a function of t. All of the answers I come up with are incorrect! Any help is greatly appreciated!!! Thanks! Answer by ankor@dixie-net.com(12684)   (Show Source): You can put this solution on YOUR website! A flag is raised while an onlooker watches from a distance of 21 feet away from the base of the flag pole (see the figure below). The flag rises vertically at a rate of 8 inches per second. Let t denote the time (in seconds) after the flag begins to rise (For simplicity, assume that when the flag begins to rise it is 0 inches above the ground). Express the distance d (in feet) between the flag and the onlooker as a function of t. : They want this in feet; convert 8" to ft : The distance from the onlooker to the flag is the hypotenuse, formed by the 21' from the base of the pole and height of the flag which would be t : d(t) = d(t) = Question 567184: -4x+5y=20 in y-intercept form and graph Answer by mananth(10539)   (Show Source): You can put this solution on YOUR website! the equation in slope intercept form is y=mx+b -4x+5y=20 Add 4x on both side -4x+4x+5y=20+4x 5y=4x+20 divide both sides by 5 y = (4/5)x+4 Question 566846: 4/7,7 WHATS THE QUADRATIC EQUATION? Answer by solver91311(12117)   (Show Source): You can put this solution on YOUR website! You need to do a couple of things before we can help you. 1. Typing in all caps is shouting. It is both annoying and rude. Stop doing it. 2. Write out the question completely. Make sure you indicate what is giving you difficulty and tell us what you have done so far. John My calculator said it, I believe it, that settles it Question 566456: factor and solve the following quadratic equation. 4x^2+10x=x^2-x+4 Answer by mananth(10539)   (Show Source): You can put this solution on YOUR website! 4x^2+10x=x^2-x+4 3x(x+4)-1(x+4)=0 (x+4)(3x-1)=0 x=-4 OR 1/3 Question 566320: Find two consecutive negative integers such that the sum of their squares is 113. Answer by ad_alta(170)   (Show Source): You can put this solution on YOUR website! Let 'n' be the smaller integer. Then n^2+(n+1)^2=113. Thus 2n^2+2n-112=0. Using the quadratic formula, we get n=-8. The two consecutive negative integers are -8 and -7. Question 566000: The directions are: Solve over the complex numbers The problems are b-3 sqrt of b -10=0 (b minus 3 times the square root of b minus 10 equal 0. and 2x-9 sqrt x + 4=0 (2x minus 9 times the square root of x plus 4) The title of the paper is the Discriminant: equations in quadratic form. Since there is no squared number in these I don't know how to begin doing them. Thank you! Answer by stanbon(48510)   (Show Source): You can put this solution on YOUR website! The problems are b-3 sqrt of b -10=0 (b minus 3 times the square root of b minus 10 equal 0. --------------------------- b - 3sqrt(b) - 10 = 0 Note: This is a quadratic with variable sqrt(b) Factor; (sqrt(b)-5)(sqrt(b)+2) = 0 sqrt(b) = 5 or sqrt(b) = -2 b = 25 or b = 4 Note: The b = 4 solution may be extraneous as sqrt(b) cannot be negative. ==================================== and 2x-9 sqrt x + 4=0 (2x minus 9 times the square root of x plus 4) 2x - 9sqrt(x) + 4 = 0 --- 2x-8sqrt(x)-sqrt(x) + 4 = 0 ---- 2sqrt(x)(sqrt(x)-4)-(sqrt(x)-4) = 0 ----- (sqrt(x)-4)(2sqrt(x)-1) = 0 sqrt(x) = 4 or sqrt(x) = 1/2 x = 16 or x = 1/4 ======================== Cheers, Stan H. =============== Question 565891: A 24-inch-wide sheet of metal is to be bent into a rectangular trough with the cross section shown in the illustration. Find the dimensions that will maximize the amount of water the trough can hold. That is, find the dimensions that will maximize the cross-sectional area.(depth and width = 24 inches) Answer by solver91311(12117)   (Show Source): You can put this solution on YOUR website! Before I even start, I want you to go back and read the question as you posted it. "...shown in the illustration." Right. This is Algebra.com -- NOT the Psychic Hot Line. First lesson: Mathematics requires using your head for something besides a hat rack. Be that as it may, I'm going to presume that you want to fold up two sides making an unknown depth (equal on both sides) on a trough with no top and a bottom that perforce measures 24 inches minus 2 times the depth. To the extent that my assumptions about the true nature of your problem are correct, begin with the concept that the area of a rectangle is given by length (or depth in this case) times width. For this problem we let represent the amount folded up or the depth of the trough, and then the width of the rectangle (the bottom of the trough must be . Multiplying length times width gives area as a function of the depth: Written in standard form: Algebra Solution Recognize that this is a quadratic polynomial function that will graph to a parabola. Since the lead coefficient is negative, the parabola must open downward, hence the vertex is a maximum. Using . Hence, fold up 6 inches to get the maximum area, giving you a 6 by 12 (24 - 12) rectangle. Calculus Solution Take the first derivative: Set the first derivative equal to zero and solve: Hence extreme point at Take the second derivative: in the domain of A. Hence the extreme point is a maximum. John My calculator said it, I believe it, that settles it Question 565872: 3(x-5)2=21 Answer by ad_alta(170)   (Show Source): You can put this solution on YOUR website! 1) 3(x-5)^2=21 2) 3x^2-30x+75=21 3) 3x^2-30x+54=0 [use the quadratic formula] 4) x=5 {+/-} sqrt(7) Question 565869: 4x2-3=9 Answer by ad_alta(170)   (Show Source): You can put this solution on YOUR website! I think you mean "4x^2-3=9." Then 4x^2-12=0 implies 4(x^2-3)=0. Thus, x={+/-}sqrt(3). Question 565570: What is the lenghts of the legs? The hypotenuse of a right triangle is 4 cm long. One leg is 1 cm longer than the other. Find the lengths of the legs. Round to the nearest tenth. Thank You =) Answer by ankor@dixie-net.com(12684)   (Show Source): You can put this solution on YOUR website! The hypotenuse of a right triangle is 4 cm long. One leg is 1 cm longer than the other. Find the lengths of the legs. Round to the nearest tenth. : Let x = one leg then (x+1) = the other leg : Remember our good friend pythag, a^2 + b^2 = c^2, so we have x^2 + (x+1)^2 = 4^2 FOIL (x+1)(x+1) x^2 + x^2 + 2x + 1 = 16 combine like terms on the left to reveal our old friend, the quadratic equation! 2x^2 + 2x + 1 - 16 = 0 2x^2 + 2x - 15 = 0 Unfortunately this will not factor, use the quadratic formula: In this equation: a=2, b=2, c=-15 : : the positive solution is all we want here x = x = 2.28 ~ 2.3 cm, one leg and 2.3 + 1 = 3.3 cm, the other leg : : Check this on your calc: enter results 3.99 ~ 4 Question 565690: Solve, x + y = 6 y = 2x - 3 I'm completely lost Daren Answer by ankor@dixie-net.com(12684)   (Show Source): You can put this solution on YOUR website! Let's see if we can find you. : Solve, x + y = 6 y = 2x - 3 : Note that the 2nd equation tell us y = (2x-3) so replace y with (2x-3) in the 1st equation x + (2x-3) = 6 3x - 3 = 6 add 3 to both sides 3x = 6 + 3 3x = 9 x = 9/3 x = 3 : Find y using y = 2x - 3, replace x with 3, and you have y = 2(3) - 3 y = 6 - 3 y = 3 : : We can check our solutions of x=3, y=3 in the 1st equation x + y = 6 3 + 3 = 6. You can check in the 2nd equation

---

Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820