Chapter 1.5, Problem 95E

To determine

Find the real solutions of equation.

Answer to Problem 95E

  $\begin{array}{l}x=\sqrt{5},x=-\sqrt{5}\\ x=2\sqrt{2},x=-2\sqrt{2}\end{array}$

Explanation of Solution

Given information:

  $x^4-13x^2+40=0$

Formula used: -

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $x^4-13x^2+40=0$

Assuming $x^2=z$ in above equation, new equation will be:

  $z^2-13z+40=0$

Now, solving the equation by Factorization method:

  $\begin{array}{l}{z}^2-13z+40=0\\ z^2-8z-5z+40=0\\ z\left(z-8\right)-5\left(z-8\right)=0\\ \left(z-8\right)\left(z-5\right)=0\end{array}$

  $z=5,z=8$

Now substituting value of $z=x^2$ in place of $z$

  $\begin{array}{l}{x}^2=5,x^2=8\\ x=\pm \sqrt{5},x=\pm 2\sqrt{2}\\ \\ \boxed{\begin{array}{l}x=\sqrt{5},x=-\sqrt{5}\\ x=2\sqrt{2},x=-2\sqrt{2}\end{array}}\end{array}$