Chapter 12.5, Problem 29WE

To determine

Solve the quadratic equation.

Answer to Problem 29WE

  $x\in \left\{-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right\}$

Explanation of Solution

Given:

The equation is :

  $4x^4+21x^2-18=0$

Calculation:

The given equation $4x^4+21x^2-18=0\Rightarrow 4{(x^2)}^2+21(x^2)-18=0$ .

Substitute $x^2=y$

So,

  $\begin{array}{l}4y^2+21y-18=0\\ \Rightarrow 4y^2+24y-3y-18=\mathrm{0.........................}[\text{factor the middle term]}\\ \Rightarrow 4y(y+6)-3(y+6)=0\\ \Rightarrow (4y-3)(y+6)=0\\ \Rightarrow 4y-3=0\text{ or }y+6=\mathrm{0.................}[\text{zero product property]}\\ \Rightarrow y=\frac{3}{4}\text{ or y}=-6\\ \Rightarrow x^2=\frac{3}{4}\text{ or }{x}^2=-\mathrm{6.........................}[y=x^2]\\ \Rightarrow x=\pm \frac{\sqrt{3}}{2}\text{ or x}=\pm \sqrt{-6}\end{array}$

Since square root takes only non-negative values, so the second values of x are not real.

Hence , $x\in \left\{-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right\}$