This is the problem in question -
Solve the equation $ a^2 - 8a + 12 = 0 $. Hence find the four values of $x$ which satisfy the equation $ (x^2 - x)^2 - 8(x^2 - x) + 12 = 0 $.
P.S. - I got the answers correctly (6,2: 3,-2 and -1,2) but the textbook i'm using doesn't explain the answers, so I want to make sure i used the right procedure.
Factoring gives $ (a-2)(a-6) = 0 $, so $ a = x^2 - x \in \{ 2, 6 \} $. Solving $$ x^2 - x = 2 $$ gives $ (x+1)(x-2) = 0 \implies x \in \{ -1, 2 \} $.
Solving $$ x^2 - x = 6 $$ gives $ (x+2)(x-3) = 0 \implies x \in \{ -2, 3 \} $.
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