Questions on Algebra: Quadratic Equation answered by real tutors!

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5t^2 – 10t = 0
--------------
t*(5t-10) = 0
Either t is zero, or (5t-10) is zero (since the product is zero).
t = 0
--------------
5t-10 = 0
5t = 10
t = 10/5 = 2

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Put the equations in the standard form:
ax^2+bx+c=0
If it isn't in this form then get it there.
Then plug the a, b, and c into the formula.
.
Ed

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any point on the graph is a solution of the equation of that graph.
example:
if your equation is y = x^2 + + x + 1, then when you look at the graph and see a point on the graph, you plot what the x value of that point is and the y value of that point is and you have a solution of the graph.
-----
if it is a quadratic equation, where the graph crosses the x axis tells you what the roots are of the equation.
these are called the x intercepts.
if the graph doesn't cross the x-axis then it doesn't have any real roots.
since a quadratic is in the form of ax^2 + bx + c,
then a, b, and c tell you a few things about the graph also.
-b/2a is the x value of the minimum / maximum point on the graph.
if the graph is pointing upwards, then -b/2a is the x value of a maximum point.
if the graph is pointing downward, then -b/2a is the x value of a minimum point.
the tails of the graph go in the opposite direction of where the head is pointing.
if x^2 is positive, then the graph is pointing downward (head down, tails up).
if x^2 is negative which can only be if is it multiplied by a negative number, such as -x^2, then the graph is pointing upward (head up, tails down).
the y intercept is found when x = 0 which is the intersection of the x axis with the y axis.
here's a graph of x^2 - 7x + 10
the roots of the equation are x = 2, and x = 5
since x^2 is positive, the graph points down (head down, tails up).
the x value of the minimum point is -b/2a = -(-7)/2 = 7/2 = 3.5
the y value of the minimum point is found by substituting x = 3.5 into the equation.
that value becomes (3.5)^2 - 7(3.5) + 10 = -2.25.
the minimum point on the graph is (3.5,-2.25) which can be seen on the graph.
if the graph is symmetric about the y value, then the axis of symmetry would be the x coordinate of the vertex which is at the minimum or maximum point on the graph.
in this graph the axis of symmetry would be x = 3.5.
the same y value would give 2 values for x which are equidistant from the axis of symmetry.
one example of symmetry is already solved where y = 0.
x = 2, and x = 5.
if the axis of symmetry is 3.5, then:
3.5 - 2 = 1.5
5 - 3.5 = 1.5
both these points are equidistant from the axis of symmetry when x = 0.
take any other 2 x value on the graph that are equidistant from 3.5 and the y values should be the same.
try 3.5 + 5 = 8.5, and 3.5 - 5 = -1.5
when x = 8.5, y = 22.75
when x = -1.5, y = 22.75
the graph is symmetric about the value of x = 3.5.

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lets call the length of the blue mat y and the width of the border x, which means the side of the entire area is y+2x
we also know that, y=(3/4)(y+2x)--->y=(3/4)y+(6/4)x--->1/4y=6/4x-->y=6x
since area of the red(28) is equal to entire area minus the blue area
we have
--->
substitute value of y which is 6x into the equation
---->

y=6x(1)=6
so the sides of the entire area is y+2x ---> meters

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A square mat has a uniform red border on all four sides. The rest of the mat is blue. The width of the blue square is three-fourths the width of the entire square. If the area colored red is 28 m squared, determine the lengths of each side of the mat.
.
Let x = length of one side of mat
and y = width of red border
.
From: "The width of the blue square is three-fourths the width of the entire square." we get equation 1:
x-2y = (3/4)x
.
From:"area colored red is 28 m squared" we get equation 2:
"area of mat" - "area of blue" = "area of red"
x^2 - (x-2y)^2 = 28
.
Solve equation 1 for y:
x-2y = (3/4)x
-2y = (3/4)x - x
-2y = -.25x
y = 0.125x
.
Substitute the above into equation 2 and solve for x:
x^2 - (x-2y)^2 = 28
x^2 - (x-2(0.125x))^2 = 28
x^2 - (x-.25x)^2 = 28
x^2 - (.75x)^2 = 28
x^2 - 0.5625x^2 = 28
0.4375x^2 = 28
x^2 = 64
x = (+-)8
.
Well, we can throw out the negative answer leaving:
x = 8 meters

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let:
W1 = width of the large square.
L1 = length of the large square.
W2 = width of the large square.
L2 = length of the large square.
-----
you are given that w = the width of the large square:
W1 = w
since this is a square, then L1 = w also.
you have:
L1 = W1 = w
-----
you are given that .75*w = the width of the small square:
W2 = .75*w
since this is a square, then L2 = .75*w also.
you have:
L2 = W2 = .75*w
-----
area of the large square is L1*W1 = A1
area of the small square is L2*W2 = A2
-----
large square is all red (assumed since you are only given that the border is red).
small square is all blue (given).
-----
if you take the small square and put it in the middle of and on top of the large square, what you will have is a blue square in the middle with a red border.
-----
that, i believe, is what causes your problem to be solved as follows:
-----
you are given that the red border is 28 square meters.
if you take the area of the large square and subtract the area of the small square you will be left with the area of the border.
so, your equation is:
A1 - A2 = 28
since A1 = L1*W1, and A2 = L2*W2, this becomes:
L1*W1 - L2*W2 = 28
-----
since L1 = w, and W1 = w, this becomes:
w*w - L2*W2 = 28
since L2 = .75*w, and W2 = .75*w, this becomes:
w*w - (.75*w)*(.75*w) = 28
simplifying:
w^2 - .5625*w^2 = 28
(1-.5625)*w^2 = 28
.4375*w^2 = 28
w^2 = 28/.4375
w^2 = 64
w = 8
-----
if w = 8, then .75*w = 6
-----
since w = 8, then
L1 = 8
W1 = 8
since .75*w = 6, then
L2 = 6
W2 = 6
-----
your original equation is:
L1*W1 - L2*W2 = 28
this becomes:
8*8 - 6*6 = 28
this becomes:
64 - 36 = 28
which becomes:
28 = 28
since this statement is true, the value for w = 8 is correct.
the lengths of each side of the mat are:
8 inches for the large square.
6 inches for the small square.
-----

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LOOKS TO ME LIKE YOU ARE RIGHT ON WITH THE FIRST TRY!!!!!!
Let W=width of entire square
Then 0.75W=width of the blue square
Area colored red would be width of entire square minus width of blue square, or:
Area colored red=W^2-0.75W^2 and we are told that this equals 28 sq m, so, our equation to solve is:
W^2-0.75W^2=28 or
0.25W^2=28 divide each side by 0.25
W^2=112 sq m take square root of each side
W=plus or minus 10.6 m disregard negative value for W
W=10.6 m
0.75W=0.75*10.6=7.95 m
CK
112-84=28
28=28
Hope this helps----ptaylor

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15 + 3y – 5x^2 – x^2y
Group terms:
(15 + 3y) – (5x^2 + x^2y)
Factor each group:
3(5 + y) – x^2(5 + y)
Factor out the (5 + y):
(5+y)(3-x^2)

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Since you are dealing with right triangles use the Pythagorean Theorem.
a^2+b^2=c^2

side a=height of rectangle=2x-1
side b=width of rectangle=3x-1
Hypotenuse=diagonal =sq rt 5

Plug in
(2x-1)^2+(3x-1)^2=5
Use the foil method to expand the expression.
4x^2-4x+1+9x^2-6x+1=5
Combine like terms to simplify

13x^2-10x+2=5
Subtract 5 from both sides

13x^2-10x-3=0
Factor
(13x+3)(x-1)=0

x=-13/3 or 1
x cannot be a negative number so the only answer is x=1
side a=height of rectangle=2x-1=2-1=1
side b=width of rectangle=3x-1=3-1=2
area of a rectangle is axb=2

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Use the distributive property to multiply the first part of the equation.
2x^2+6x=x+25
subtract x from both sides
2x^2+5x=25
Subtract 25 from both sides
2x^2+5x-25=0
this is a quadratic equation

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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=225 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 2.5, -5. Here's your graph:

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16x^4 - 40x^2 + 9
= 16x^4 - 36x^2 - 4x^2 + 9
= 4x^2(4x^2 - 9) - 1(4x^2 - 9)
= (4x^2 - 9)(4x^2 - 1)
= (2x + 3)(2x - 3)(2x + 1)(2x - 1)

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9b^3-6b^2
= 3b^2(3b-2)

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(x + 4)(x + 3) = 6
x^2 + 3x + 4x + 12 = 6
x^2 + 7x + 12 - 6 = 6 - 6
x^2 + 7x + 6 = 0
x^2 + 6x + x + 6 = 0
x(x + 6) + 1(x + 6) = 0
(x + 1)(x + 6) = 0
x = -1,-6

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(x + 4) (x + 3) = 6
First, expand the left with FOIL:
x^2 + 3x + 4x + 12 = 6
x^2 + 7x + 12 = 6
x^2 + 7x + 6 = 0
Now, factoring the left side:
(x+1)(x+6) = 0
.
x = {-1, -6}

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If you use Google on the internet you will find a myriad of examples.
Cheers,
Stan H.

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Start with the given equation.


Notice we have a quadratic equation 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 .


Take the square root of to get .


or Break up the expression.


or Combine like terms.


or Simplify.


So the answers are or

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Start with the given equation.


Get all terms to the left side.


Combine like terms.


Notice we have a quadratic equation 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 answers are or


which approximate to or

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Start with the given equation.


Get all terms to the left side.


Combine like terms.


Notice we have a quadratic equation 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


Subtract from to get


Multiply and to get .


or Break up the expression.


So the answers are or


which approximate to or

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Start with the given equation.


Take the square root of both sides.


or Break up the "plus/minus" to form two equations.


or Take the square root of to get .


or Subtract from both sides.


or Combine like terms.


--------------------------------------


Answer:


So the solutions are or .

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If you are looking at a graph of a quadratic equation, how do
you determine where the solutions are?
----------------
If it's the usual y = a quadratic in x, it's where it crosses the x-axis.
For example:

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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=49 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 4, -3. Here's your graph:

If there are no real solutions, it won't cross the x-axis.

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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 : . The discriminant -36 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -36 is + or - . The solution is , or Here's your graph:

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I'll do the first (which hopefully will help you do the second)


# 1




Start with the given equation


Add to both sides


Complete the square for the "x" terms. Note: Let me know if you need help completing the square.


Complete the square for the "y" terms


Combine like terms


Add to both sides


Combine like terms




-----------


Notice how the equation is now in the form . This means that this conic section is a circle where (h,k) is the center and is the radius.

So the circle has these properties:

Center: (4,-2)

Radius:

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Start with the given equation.


Multiply every term on both sides by the LCD . Doing this will eliminate all of the fractions.


Multiply and simplify


FOIL


Distribute


Combine like terms.


Add 20 to both sides


Combine like terms


Divide both sides by 5


or Take the square root of both sides.


or Simplify the square root.



============================================

Answer:

So the solutions are or

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( x2 – 2x )( x + 3) = -2x ( x + 1)
x3 + 3x2 - 2x2 -6x = -2x2 - 2x
x3 + 3x2 - 4x = 0

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I am thinking of three consecutive negative numbers. If I multiply the first with the second and then subtract three times the third, the result is 57. What are the numbers?
solution:
Let the three numbers be x, x+1 and x+2
x(x+1)-3(x+2) = 57
x^2+x-3x-6=57
x^2-2x-63=0
x^2-9x+7x-63=0
(x+7)(x-9)=0
Hence x = -7 or x=9
But the three numbers shd be negative.So x= -7
Hence the three consecutive numbers are -7,-8,-9

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( x2 – 2x )( x + 3) = -2x ( x + 1)
x(x-2)(x+3)=-2x(x+1)
x(x-2)(x+3)+2x(x+1)=0
x{(x-2)(x+3)+2(x+1)}=0
x=0
or
(x-2)(x+3)+2(x+1) =0
x^2+x-6+2x+2 = 0
x^2+3x-4=0
x^2+4x-x-4=0
(x-1)(x+4)=0
Hence the solutions are
x=0, x=1 or x=-4

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The number of people infected after t days is P = 200 + 5t - t2. Find the number of days until the number of people infected is 176.
solution:
Let the number of people infected afetr t days be P
P = 200 + 5t - t^2
we haev to find t when P = 176
we have to solve for t
200+ 5t - t^2 = 176
t^2-5t-24=0
t^2-8t+3t-24=0
(t-3)(t-8)=0
Hence t=3 or t=8

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6b^4-18b^3-60b^2
= 6b^2(b^2-3b-10)
=6b^2(b^2-5b+2b-10)
=6b^2(b-5)(b-2)

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Zarig took part in a 26 mile road race.
a. he ran the first 15 miles at an average speed of x mph. he ran the last 11 miles at an average speed of (x-2)mph. write down an expression,in terms of x, for the time he took to complete the 26 mile race?
-------
time = distance/rate
Total time = 15/x + 11/(x-2)
-----------------------------------
b.zarig took 4 hours to complete the race. using your answer to part a, form an equation in terms of x.
4 = 15/x + 11/(x-2)
------------------------
c. i) simplify your equation and show that it can be written as
2x^2-17x+15=0
---------
4 = 15/x + 11/(x-2)
4x(x-2) = 15(x-2) + 11x
4x^2-8x = 15x-30 + 11x
4x^2-34x-30 = 0
2x^2-17x-15 = 0
---------------------------
ii) solve the equation and obtain zarig's average speed over the first 15 miles of the race.
x = [17 +- sqrt(17^2-4*2*-15)]/4
x = [17 +- sqrt(409)]/4
x = [17 +- 20.2237..]/4
Positive solution:
x = 9.306 mph (average speed over the 1st 15 miles)
===================
Cheers,
Stan H.

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Multiply out each side of the equation
3x^2+9x=10x+2
subtract 10x from both sides
3x^2-x=2
subtract 2 from both sides
3x^2-x-2=0
Factor the quadratic equation.

(3x+1)(x-2)=0
x=-1/3 or x=2

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50 is what pecent of 80?
x * 80/100 = 50
=> x = 500/8 = 62.5
82 of what number is 16?
x/100 * 82 = 16
=> x = 1600/82=800/41 = 19.51

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A firm can sell X units of a product at a price of P cents per unit, where P = 503 - X
The total costs of production is given by the function T(X) = 500 + 2X.
# Find the number of units to be sold to achieve maximum profit and find the maximum profit.
# What price per unit would the firm be charging at the maximum profit output level?
--------------
Revenue = price*# of units sold
Revenue = (503-x)(X) = 503x - x^2
-----------------
Cost = 500+2x
-----------------
Profit = [Revenue - Cost] = (503x-x^2)-(500+2x) = -500+501x-x^2
--------------
Maximum profit occurs when x = -b/2a = -501/(2*-1) = 250.5
-------------------------------------
Maximum profit will be -500 + 501*250.5-250.5^2 = $62,250.25
--------------------------
Price when profit is maximum = 503-250.5 = $252.50
============================================================
Cheers,
Stan H.

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(3x+1)(x-1)= -3x
Start by multiplying both factors on the left side using FOIL.
3x^2 - 3x + x - 1 = -3x
3x^2 -2x - 1 + 3x = 0
3x^2 + x - 1 = 0
Use the quadratic formula to factor and find x.
Can you take it from here?

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Start with the left side of the equation.


Take half of the coefficient to get . In other words, .


Now square to get . In other words,


Now add and subtract . Make sure to place this after the "s" term. Notice how . So the expression is not changed.


Group the first three terms.


Factor to get .


Combine like terms.


So after completing the square, transforms to . So .


So is equivalent to .

-----------------------------------------------


Start with the completed square set to 0.


Add to both sides.


Combine like terms.


Take the square root of both sides.


or Break up the "plus/minus" to form two equations.


or Take the square root of to get . Note: so


or Add to both sides.


or Combine the fractions


or Combine the numerators


or Reduce.


--------------------------------------


Answer:


So the solutions are or .

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The annual yield per lemon tree is fairly constant at 320 pounds when the number
of trees per acre is 50 or fewer. For each additional tree over 50, the annual
yield per tree for all trees on the acre decreases by 4 pounds due to
overcrowding. Find the number of trees that should be planted on an acre to
produce the maximum yield.
;
Let x = no. of trees, over 50, planted
Then y = total yield per acre
:
total yield = No.of trees * lbs of lemons per tree
:
y = (50+x) * (320 - 4x)
FOIL
y = 16000 - 200x + 320x - 4x^2
The quadratic equation
y = -4x^2 + 120x + 16000
:
The axis of symmetry will be the value of x for max y
The formula for that: x =
In this problem: a=-4; b=120
x =
x =
x = +15 trees over 50 (65) gives max yield per acre
:
How many pounds is the maximum yield?
:
you can find that, substitute 15 for x in the quadratic equation

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the walk is on all four sides of the pool, so the dimensions of the pool are 20-2w and 22-2w
__ where w is the width of the walk

(20-2w)(22-2w)=323 __ 440-84w+4w^2=323 __ 4w^2-84w+117=0

use quadratic formula to find w




answer below


w=1.5

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P=x^2-13x-80
How many watches were sold on a day when there was a $50 loss?
.
This means P is -50.
Plug it in and solve for x:
P=x^2-13x-80
-50=x^2-13x-80
0=x^2-13x-30
factoring:
0=(x-15)(x+2)
.
x = {-2, 15}
.
We can toss out the negative answer so we get:
x = 15 watches

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5x/4 + 1/2=x-1/2
re-writing:
, combine similar terms

, cross multiply

Check by going back:





Thank you,
Jojo

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Your method is called "completing the square"...
This site explains it quite nicely:
http://www.purplemath.com/modules/sqrquad.htm
.
x^2-2x-13=0
x^2-2x = 13
Take half the 'b' coefficient and square it:[(1/2)(-2)]^2 = [-1]^1 = 1
x^2-2x+1 = 13+1 (since you added 1 to left, do so on the right - for balance)
(x-1)^2 = 14
x-1 = sqrt(14)
x = 1(+-)sqrt(14)
.
That's 1 "plus or minus" square root of 14.
.
****************************
4x^2-4x+3=0
4x^2-4x = -3
factor the 4 on the left:
4(x^2-x) = -3
(x^2-x) = -3/4
(x^2-x+(1/4)) = -3/4 + 1/4
(x-(1/2))^2 = -2/4
x-(1/2) = sqrt(-2/4)
x = (1/2)(+-)sqrt(-1/2)
.
Since the term inside the sqrt is negative -- we have no real solutions -- rather, we have two imaginary solutions.
.
You could see the same thing using the "quadratic equation" below:

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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 : . The discriminant -32 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -32 is + or - . The solution is Here's your graph:

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5x+3=23
23-3=20
__20___ = 4
5x


check: 5(4)+3=23
20 + 3 = 23

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subtracting 4-8x __ x^2+8x-4=0 __ this is not factorable

adding 4 __ x^+8x=4 __ completing the square by adding (8/2)^2 __ x^2+8x+16=20

taking square root __ x+4=±sqrt(20) __ x=-4 ± 2(sqrt(5))

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the reciprical of x is 1/x.
the reciprical of a/b is b/a
.
let x represent the "certain positive number"
Therefore, the reciprical of x is 1/x and 1/x is one less than twice x so:

.
Solve for x
multiply both sides by x (remember to put parenthesis around the right side so you do not make an error in multiplying).

reduce on the left and distribute on the right side

move all terms to the same side

.
From here you can either plug this into the Quadratic equation or you can complete the square.
.

.

.

.

.
and
.
I hope this helps!
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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4x^2 + 6.4 x + k-------------------------------------eq1
Let's first assume that the solution is the following and then solve for b:
(2x+b)^2 expand using FOIL;
4x^2+2bx+2bx+b^2=
4x^2+4bx+b^2--------------------Now this equation has to be identical to eq1 in all respects, so:
4b=6.4;
b=6.4/4=3.2/2----eq2
and
b^2=k----------eq3
substitute eq2 into eq3
(3.2/2)^2=k or
k=(3.2/2)^2
substitute k=(3.2/2)^2 into eq1
4x^2+6.4x+(3.2/2)^2=(2x+3.2/2)^2
CK
(2x+3.2/2)^2=4x^2 +(6.4/2)x +(6.4/2)x +(3.2/2)^2=
4x^2+6.4x+(3.2/2)^2

Hope this helps---ptaylor

You can put this solution on YOUR website!
x varies directly as the square of s and inversely as t. H
x =
:
How does x change when s is doubled?
x = = ; x increase 4 times
:
:
When both s and t are doubled?
x = = = : We can say that x is doubled

You can put this solution on YOUR website!
No.
+or-=-or+

Hope this helps---ptaylor

You can put this solution on YOUR website!
9n2=-5+7n
9n^2-7n+5=0

n=(7+-sqrt[-7^2-4*9*5])/2*9
n=(7+-sqrt[49-180])/18
n=(7+-sqrt-131)/18
n=(7+-11.4455i)/18
n=7/18+-11.4455i/18
n=.388888+-.63586i answers.