Prove equation is -ve for all real values

**mathguy80** · May 9th 2011, 05:18 AM

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!

**red_dog** · May 9th 2011, 12:26 PM

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

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

**mathguy80** · May 9th 2011, 07:08 PM

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!

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