Solve the equation by making an appropriate substitution.
-$\left( y - \frac { 12 } { y } \right) ^ { 2 } - 3 \left( y - \frac { 12 } { y } \right) - 4 = 0$
A) $\{ - 4 , - 2,3,6 \}$
B) $\{ - 1,4 \}$
C) $\{ - 4,3 \}$
D) no solution

Question

Quizplus · Accepted Answer

Let $u = y - \frac{12}{y}$. The equation becomes $u^2 - 3u - 4 = 0$, which factors to $(u - 4)(u + 1) = 0$, giving $u = 4$ or $u = -1$. Substituting back, for $u = 4$, $y - \frac{12}{y} = 4$ leads to $y^2 - 4y - 12 = 0$, which factors to $(y - 6)(y + 2) = 0$, giving $y = 6$ or $y = -2$. For $u = -1$, $y - \frac{12}{y} = -1$ leads to $y^2 + y - 12 = 0$, which factors to $(y - 3)(y + 4) = 0$, giving $y = 3$ or $y = -4$. Thus, the solution set is $\{-4, -2, 3, 6\}$.

Solve the Equation by Making an Appropriate Substitution (y−12y)2−3(y−12y)−4=0\left( y - \frac { 12 } { y } \right) ^ { 2 } - 3 \left( y - \frac { 12 } { y } \right) - 4 = 0(y−y12​)2−3(y−y12​)−4=0

Solve the Equation by Making an Appropriate Substitution $\left( y - \frac { 12 } { y } \right) ^ { 2 } - 3 \left( y - \frac { 12 } { y } \right) - 4 = 0$