Solve the equation for y. (y -3)^2 = 4y -12

Answer 1

First, let's expand the (y-3)^2

Then we will combine like terms and set the equation equal to zero.

Lastly we will factor and use the zero product rule to solve.

(y-3)^2 = 4y-12

(y-3)(y-3)=4y-12

y^2 -6y +9 = 4y - 12

y^2-10y +21= 0

Think of two numbers whose product (answer when we multiply) is 21 and whose sum (answer when we add) is -10. How about -7 and -3?

Okay we are ready to factor.

(y-7)(y-3) = 0 Therefore, either y-7 = 0 or y-3 = 0 So y = 7, 3

Hope that makes sense!

Answer 2

There are two ways to solve this problem. One is to expand the left-hand side and find the roots of the quadratic. The more satisfactory is to first factor the right side.

(y-3)^2 = 4y - 12

(y-3)(y-3) = 4(y-3)

(y-3)(y-3) - 4(y-3) = 0

(y- 3 - 4)(y-3) = 0

(y-7)(y-3) = 0

y = 3, 7