Find all real solutions. x⁴ + x² - 4 = 0 A) $x = \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }$ B) $x = \sqrt { \frac { 1 - \sqrt { 13 } } { 2 } }$ C) $x = \pm \sqrt { \frac { 1 - \sqrt { 13 } } { 2 } }$ D) $x = \pm \sqrt { \frac { - 1 \pm \sqrt { 17 } } { 2 } }$ E) $x = \pm \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }$

Question

Quizplus · Accepted Answer

The equation can be solved by substituting $ y = x^2 $, leading to $ y^2 + y - 4 = 0 $. Solving this quadratic equation gives $ y = \frac{-1 \pm \sqrt{17}}{2} $. Since $ y = x^2 $, the real solutions for $ x $ are $ x = \pm \sqrt{\frac{-1 + \sqrt{17}}{2}} $.

Find All Real Solutions x=−1+172x = \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }x=2−1+17​​​

Find All Real Solutions $x = \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }$