Answer This Please (1 Viewer)

[Andy005]

One of the roots of the cubic x^3+ax^2+36x-36=0 is the product of the other two roots. Find the value of a and hence find all the roots.

 

[si2136]

Let roots be XY, X, Y.

Sum of Roots = -b/a = -a = XY + X + Y
Sum of Product of Roots = c/a = X^2 Y + XY^2 + XY = 36
Prod. of Roots = -d/a = X^2 Y^2 = 36

X^2 Y + XY^2 + XY = XY(X+Y+1) = 36

XY = 6

Therefore X+Y+1 = 6

X+Y = 5

Therefore 6 + 5 = -a

Therefore = a = 11

 

[Andy005]

I don't understand how you got to a=11 and I think it's wrong because x^3+11x^2+36x-36=0 has an approximate solution x= 0.79365 , and this solution doesn't satisfy the question.
Can you please explain in more detail because I'm not exactly sure how to start this.

Thanks

 

[si2136]

I don't understand how you got to a=11 and I think it's wrong because x^3+11x^2+36x-36=0 has an approximate solution x= 0.79365 , and this solution doesn't satisfy the question.
Can you please explain in more detail because I'm not exactly sure how to start this.

Thanks

OOPS - Made a typo on the last step, should be a = -11.

Full working out here: View attachment 33866

 

[Andy005]

I have trouble understanding what roots to use for certain questions, do you have any advice?

 

[si2136]

I have trouble understanding what roots to use for certain questions, do you have any advice?

Usually they'll have one root change and the otherones fixed. For example, let one root be the sum of the other two. Then roots be a+b, a and b.