Chapter 9.6, Problem 16PPS

To determine

The solutions of the equation $4x^4-33x^2+50=0$ using quadratic formula

Answer to Problem 16PPS

  $x=\pm \frac{5}{2},\text{ and }x=\pm \sqrt{2}$

Explanation of Solution

Given:

The equation,

  $4x^4-33x^2+50=0$

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

In order to solve the given equation $4x^4-33x^2+50=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=4,b=-33,\text{ and }c=50$

Now, above quadratic equation $4y^2-33y+50=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-33\right)\pm \sqrt{{\left(-33\right)}^2-4\left(4\right)\left(50\right)}}{2\left(4\right)}\\ =\frac{33\pm \sqrt{1089-800}}{8}\\ =\frac{33\pm \sqrt{289}}{8}\\ =\frac{33\pm 17}{8}\\ =\frac{33+17}{8},\text{ and }y=\frac{33-17}{8}\\ \Rightarrow y=\frac{50}{8},\text{ and }y=\frac{16}{8}\\ \Rightarrow y=\frac{25}{4},\text{ and }y=2\end{array}$

Thus, the solutions of the given question are, $x=\pm \frac{5}{2},\text{ and }x=\pm \sqrt{2}$ .