Chapter 9.7, Problem 30SR

To determine

The solutions of the equation $2x^2=-2x+11$ using quadratic formula

Answer to Problem 30SR

  $x=\frac{1}{2}+\frac{\sqrt{23}}{2},\text{ and }x=\frac{1}{2}-\frac{\sqrt{23}}{2}$

Explanation of Solution

Given:

The equation,

  $2x^2=-2x+11$

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 $2x^2+2x-11=0$ , 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

  $a=2,b=2,\text{ and }c=-11$

Now, above quadratic equation $2y^2+2y-11=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(2\right)\pm \sqrt{{\left(2\right)}^2-4\left(2\right)\left(-11\right)}}{2\left(2\right)}\\ =\frac{-2\pm \sqrt{4+88}}{4}\\ =\frac{-2\pm \sqrt{92}}{4}\\ =\frac{-2\pm \sqrt{23\times 4}}{4}\\ =\frac{-2\pm 2\sqrt{23}}{4}\\ x=\frac{-2+2\sqrt{23}}{4},\text{ and }x=\frac{-2-2\sqrt{23}}{4}\\ x=\frac{2\left(1+\sqrt{23}\right)}{4},\text{ and }x=\frac{2\left(1-\sqrt{23}\right)}{4}\\ x=\frac{\left(1+\sqrt{23}\right)}{2},\text{ and }x=\frac{\left(1-\sqrt{23}\right)}{2}\\ \Rightarrow x=\frac{1}{2}+\frac{\sqrt{23}}{2},\text{ and }x=\frac{1}{2}-\frac{\sqrt{23}}{2}\end{array}$

Thus, the solutions of the given question are, $x=\frac{1}{2}+\frac{\sqrt{23}}{2},\text{ and }x=\frac{1}{2}-\frac{\sqrt{23}}{2}$ .