Tutors Answer Your Questions about Quadratic Equations (FREE)
Question 353827: A man deposited $2000 in a savings bank paying him a certain interest. At the end of the year he withdraw $150. At the end of the second year the amount dew to him was $1947.50. What was the rate of interest?
Click here to see answer by mananth(16946)  |
Question 353840: Leah and Gellie can finish a project in 4 hours if they worked together. If it would take Leah 6 hours more than Gellie to finish the project alone, how long would Leah need to finish the project working alone? (Show your solution using Quadratic Equation)
Click here to see answer by stanbon(75887) |
Question 353894: I have a hard exponent quadratic question.
5x^2/3 + 2x^4/3 -13 =0
My first inclination was to rearrange the terms to...
2x^4/3 + 5x^2/3 -13 =0
This puts it in the right order for substitution / y= x^2/3
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.
Thanks for your help.
Neil
Click here to see answer by Earlsdon(6294) |
Question 353894: I have a hard exponent quadratic question.
5x^2/3 + 2x^4/3 -13 =0
My first inclination was to rearrange the terms to...
2x^4/3 + 5x^2/3 -13 =0
This puts it in the right order for substitution / y= x^2/3
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.
Thanks for your help.
Neil
Click here to see answer by Alan3354(69443)  |
Question 353894: I have a hard exponent quadratic question.
5x^2/3 + 2x^4/3 -13 =0
My first inclination was to rearrange the terms to...
2x^4/3 + 5x^2/3 -13 =0
This puts it in the right order for substitution / y= x^2/3
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.
Thanks for your help.
Neil
Click here to see answer by ewatrrr(24785)  |
Question 353887: I have a problem that are driving me crazy. The problem has a quadratic solution, but for the life of me I can't figure it out. Please help!
5(3x-4)^6 + 7(3x-4)^3 -23 =0
I set it up by substitution / y= (3x-4)^3
This gives me the form 5y^2 + 7y -13 =0
My problem is not knowing how to "unfold" the problem after I've done the alternate equation.
Thanks for your help
Click here to see answer by jim_thompson5910(35256) |
Question 353963: I'm a little confused on how to work this one...
4x^4 - 12x^2 - 19 = 0
I use subtitution of y = x^2 giving me
4y^2 - 12y - 19 = 0
My answers are y = 4.145751311 and -1.145751311
Now I'm not sure what to do from here.
Thanks a lot for your help!
Neil
Click here to see answer by jim_thompson5910(35256) |
Question 353969: Earlier I asked for help to solve this.
5(3x-4)^6 + 7(3x-4)^3 -23 = 0
The tutor sent back an answer I don't know how to punch into my calculator to get the real number. The form they sent was not in whole number form, but in a form I don't know how to even punch in my calculator yet. Any further help would be great!
Neil
Click here to see answer by jim_thompson5910(35256) |
Question 354305: I need to factor 25c^2+1-10c
This is what I've tried, but I'm just lost.
25c^2-10c+1=0
25c^2-5c-5c+1
(25c^2+1)(-5c-5c)
(5c+1)(5c+1)-10c
I don't think this is right. Please help me find what I'm doing wrong.
Click here to see answer by Earlsdon(6294) |
Question 354318: I am supposed to find all #s for which rational expression is unidentified.
(q^3-6q)/(q^2-9)
I got q(q^2-6)/(q+3)(q-3). Is this right? Does it mean the solutions that are unidentified are 3 and -3?? Because they would make the denominator zero?
Click here to see answer by scott8148(6628)  |
Question 354730: We have been working on factoring trinomials and the problem I don't know how to solve is (x^2+4x-4)^2=64. I think what we're supposed to do is divide the exponent off both sides but I'm not sure. Were I get stuck is you're left with x^2+4x-4 on one side and that doesn't facter into two binomials. HELP ME!
Click here to see answer by scott8148(6628)  |
Question 354730: We have been working on factoring trinomials and the problem I don't know how to solve is (x^2+4x-4)^2=64. I think what we're supposed to do is divide the exponent off both sides but I'm not sure. Were I get stuck is you're left with x^2+4x-4 on one side and that doesn't facter into two binomials. HELP ME!
Click here to see answer by Edwin McCravy(20054)  |
Question 355000: I need to solve the equation by completing the square. The following is my work but I get the wrong answer. Please let me know where I went wrong.
x^2 - 6x +3 = 0
x^2 -6x = -3
x^2 -6x -3^2 = -3-3^2
x^2 - 6x = -12
(x-6)^2 = sqrt -12
x-6 = sqrt -12
x=6+sqrt -12 and x = 6 - sqrt-12
Click here to see answer by jim_thompson5910(35256) |
Question 355007: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= . I must write the quadratic equation into vertex form and find the vertex.
What I did was factor out the 5 from the equation. So the result would be y= . To complete the perfect square trinomial, I made the equation y= or adding 1 to both sides. Moving the constants to one side the equation came out to be y= leaving the vertex to be (-1,6).
Is this correct?
Click here to see answer by jim_thompson5910(35256) |
Question 355007: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= . I must write the quadratic equation into vertex form and find the vertex.
What I did was factor out the 5 from the equation. So the result would be y= . To complete the perfect square trinomial, I made the equation y= or adding 1 to both sides. Moving the constants to one side the equation came out to be y= leaving the vertex to be (-1,6).
Is this correct?
Click here to see answer by stanbon(75887) |
Question 355495: Hi, I'm being asked to solve for x by completing the square:
-2x^2 + 12x - 6 = 0
-2x^2 + 12x = 6 < I moved the 6 to the other side
x^2 - 6x = -3 < I divided everything by -2 to get the x^2 by itself
x^2 - 6x + 9 = 6 < I took half of the coefficient 6 (6/2 = -3), squared
it (9), and added it to both sides of the equation
This is where I get stuck. I'm supposed to write the left side in squared form, but because there is a plus and minus, it wouldn't be just one factor squared. Would I write this as (x-3)(x+3)? If so, how would I take the square root of that?
Thanks for any guidance you can provide!
Click here to see answer by scott8148(6628)  |
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