# Thread: Prove equation is -ve for all real values

1. ## Prove equation is -ve for all real values

Hi Guys,

This problem from quadratic equations exercises that I am doing. Not sure how to proceed. Your help is much appreciated.

Prove that the expression $\displaystyle (24 - x)(x - 8) - k$ is negative for all real values of x provided that k is greater than 64. Hence show that the expression $\displaystyle (6 + y)(4 - y)(y + 4)(y - 2) - 65$ is negative for all real values of y.

The first part I solved by completing the square to get $\displaystyle -(x + 16)^2 + (64 - k)$. So -ve for $\displaystyle k > 64$.

Having trouble with the rest of the question. I tried simplifying but I end with an equation for order 4. What am I doing wrong?

2. $\displaystyle (6+y)(4-y)(4+y)(y-2)-65=(24-y^2-2y)(y^2+2y-8)-65$
Noe substitute $\displaystyle y^2+2y$ with $\displaystyle x$ and you get $\displaystyle (24-x)(x-8)-65$, then apply the first part of the problem.
$\displaystyle (6+y)(4-y)(4+y)(y-2)-65=(24-y^2-2y)(y^2+2y-8)-65$
Noe substitute $\displaystyle y^2+2y$ with $\displaystyle x$ and you get $\displaystyle (24-x)(x-8)-65$, then apply the first part of the problem.