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

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

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?

Thanks for your help!

2. $(6+y)(4-y)(4+y)(y-2)-65=(24-y^2-2y)(y^2+2y-8)-65$

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

3. Originally Posted by red_dog
$(6+y)(4-y)(4+y)(y-2)-65=(24-y^2-2y)(y^2+2y-8)-65$

Noe substitute $y^2+2y$ with $x$ and you get $(24-x)(x-8)-65$, then apply the first part of the problem.
Thanks @red_dog, seems so simple now!