Solve the equation by making an appropriate substitution.
-$$\left( x ^ { 2 } - 3 x \right) ^ { 2 } - 14 \left( x ^ { 2 } - 3 x \right) + 40 = 0$$
A) $\{ 10,4 \}$
B) $\{ - 2 , - 1,10,4,5,4 \}$
C) $\{ - 2 , - 1,5,4 \}$
D) $\{ 5,4 \}$

Question

Quizplus · Accepted Answer

Let $u = x^2 - 3x$. The equation becomes $u^2 - 14u + 40 = 0$. Factoring or using the quadratic formula gives $u = 10$ or $u = 4$. Substituting back for $x^2 - 3x$, we solve $x^2 - 3x - 10 = 0$ and $x^2 - 3x - 4 = 0$, yielding solutions $x = -2, 5$ and $x = -1, 4$, respectively.

Solve the Equation by Making an Appropriate Substitution (x2−3x)2−14(x2−3x)+40=0\left( x ^ { 2 } - 3 x \right) ^ { 2 } - 14 \left( x ^ { 2 } - 3 x \right) + 40 = 0(x2−3x)2−14(x2−3x)+40=0

Solve the Equation by Making an Appropriate Substitution $\left( x ^ { 2 } - 3 x \right) ^ { 2 } - 14 \left( x ^ { 2 } - 3 x \right) + 40 = 0$