Chapter 6, Problem 1CR

To determine

To calculate: The solution of the equation $2x^2+x-1=0$

Answer to Problem 1CR

Solution:

The real solutions of the equation $2x^2+x-1=0$ are $x=\frac{1}{2}$ or $x=-1$

Explanation of Solution

Given information:

The equation $2x^2+x-1=0$

Formula used:

Solutions of the quadratic equation in the standard form $f\left(x\right)=a{x}^2+bx+c$ are given by the quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Comparing $2x^2+x-1=0$ with $a{x}^2+bx+c=0$

  $a=2,b=1,\text{and}c=-1$ .

Substitute the values in the quadratic formula and simplify:

  $\Rightarrow x=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(2\right)\left(-1\right)}}{2\left(2\right)}$

  $\Rightarrow x=\frac{-1\pm \sqrt{1+8}}{4}$

  $\Rightarrow x=\frac{-1\pm \sqrt{9}}{4}$

  $\Rightarrow x=\frac{-1\pm 3}{4}$

  $\Rightarrow x=\frac{-1+3}{4}\text{or}x=\frac{-1-3}{4}$ .

  $\Rightarrow x=\frac{2}{4}\text{or}x=\frac{-4}{4}$

  $\Rightarrow x=\frac{1}{2}\text{or}x=-1$

Therefore, the real solutions of the quadratic equation are: $x=\frac{1}{2}\text{or}x=-1$