Find all values of x satisfying the given conditions.
-$y = \left( x - \frac { 12 } { x } \right) ^ { 2 } - 3 \left( x - \frac { 12 } { x } \right)$ and $y = 4$
A) $- 4,3$
B) $- 1,4$
C) $- 4 , - 2,3,6$
D) No solution

Question

Quizplus · Accepted Answer

To find the values of $x$ that satisfy the given conditions, we set $y = 4$ and solve the equation:$$4 = \left( x - \frac { 12 } { x } \right) ^ { 2 } - 3 \left( x - \frac { 12 } { x } \right)$$Let $z = x - \frac{12}{x}$. Then the equation becomes:$$4 = z^2 - 3z$$Rearranging gives:$$z^2 - 3z - 4 = 0$$Factoring or using the quadratic formula gives:$$(z - 4)(z + 1) = 0$$So, $z = 4$ or $z = -1$.For $z = 4$:$$x - \frac{12}{x} = 4$$$$x^2 - 4x - 12 = 0$$Factoring or using the quadratic formula gives:$$(x - 6)(x + 2) = 0$$So, $x = 6$ or $x = -2$.For $z = -1$:$$x - \frac{12}{x} = -1$$$$x^2 + x - 12 = 0$$Factoring or using the quadratic formula gives:$$(x - 3)(x + 4) = 0$$So, $x = 3$ or $x = -4$.Therefore, the values of $x$ that satisfy the given conditions are $-4, -2, 3, 6$.

Find All Values of X Satisfying the Given Conditions y=(x−12x)2−3(x−12x)y = \left( x - \frac { 12 } { x } \right) ^ { 2 } - 3 \left( x - \frac { 12 } { x } \right)y=(x−x12​)2−3(x−x12​)

Find All Values of X Satisfying the Given Conditions $y = \left( x - \frac { 12 } { x } \right) ^ { 2 } - 3 \left( x - \frac { 12 } { x } \right)$