SOLUTION: Find the quadratic equation with roots which are the squares of the roots of x^2 -5x -3 =0

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Question 285937: Find the quadratic equation with roots which are the squares of the roots of
x^2 -5x -3 =0
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 -5x -3 =0
x^2 -5x =3
(5/2)^2=25/4
3=12/4
x^2-5x+25/4=3+25/4
(x-5/2)^2=37/4
x=5/2-sqrt(37/4)
x=5/2+sqrt(37/4)
c=1/2(5-sqrt(37))
d=1/2(5+sqrt(37))
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A1%2A-3=37.

Discriminant d=37 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+37+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+37+%29%29%2F2%5C1+=+5.54138126514911
x%5B2%5D+=+%28-%28-5%29-sqrt%28+37+%29%29%2F2%5C1+=+-0.54138126514911

Quadratic expression 1x%5E2%2B-5x%2B-3 can be factored:
1x%5E2%2B-5x%2B-3+=+1%28x-5.54138126514911%29%2A%28x--0.54138126514911%29
Again, the answer is: 5.54138126514911, -0.54138126514911. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-3+%29


let's call the roots c and d
so the new equation will have x^2 and y^2 as there roots
(a-c^2)*(b-d^2)=0
ab-bc^2-ad^2+c^2d^2=0
plug these in .
c=1/2(5-sqrt(37))
d=1/2(5+sqrt(37))