Solve the equation of quadratic form. $x - 4 \sqrt { x } - 12 = 0$
A) $x = - 4 ; x = 36$
B) $x = - 2 ; x = 6$
C) x=36
D) $x = 2 ; x = - 6$
E) $x = \sqrt { 2 } i ; x = \sqrt { 6 }$

Question

Quizplus · Accepted Answer

The equation can be rewritten as $(\sqrt{x})^2 - 4\sqrt{x} - 12 = 0$. Let $y = \sqrt{x}$, then $y^2 - 4y - 12 = 0$. Solving this quadratic equation gives $y = 6$ or $y = -2$. Since $y = \sqrt{x}$, $y$ must be non-negative, so $y = 6$. Thus, $x = y^2 = 36$.

Solve the Equation of Quadratic Form x−4x−12=0x - 4 \sqrt { x } - 12 = 0x−4x​−12=0

Solve the Equation of Quadratic Form $x - 4 \sqrt { x } - 12 = 0$