SAT math hard problem

[gritty gale]

(x-k)^(2)=(k-5a)(x-k) In the given equation, a and k are constants, where k>5a. The sum of the solutions to the equation is 3k+33. What is the value of a ? Is a =0?

[ripe merlin]

Foil it out and get all the terms on one side of the expression

So that you have a quadratic equal to 0

The constants next to the x terms are the sum of the roots of the quadratic

[gritty gale]

does a=0

[lethal phoenix]

(x-k)² = (k-5a)(x-k)
x²-2kx+a² = kx-k²-5ax+5ak
x²-2kx-kx+5ax+a²+k²-5ak = 0
x²-3kx+5ax+a²+k²-5ak
x²+5ax-3kx+a²+k²-5ak = 0
x²+(5a-3k)x+(a²+k²-5ak) = 0
using Vieta's formula, we know that x1+x2 = -b/a, therefore:
3k+33 = -(5a-3k)/1
3k+33 = -(5a-3k)
3k+33 = -5a+3k
33 = -5a
a = -33/5
so a ≠ 0

[gloomy kelp]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[gritty gale]

just to confirm

i think you might have made a mistake at x²-2kx+a² = kx-k²-5ax+5ak. where is a^2 coming from? Shouldnt it be k^2? But it doesnt matter too much since you still get -33/5