Chapter 13, Problem 1CR

To determine

To calculate: The solution of the equation $3x^2-2x=-1$.

Answer to Problem 1CR

Solution:

The solution set of the equation $3x^2-2x=-1$ is $\left\{\frac{1-\sqrt{2i}}{3},\frac{1+\sqrt{2i}}{3}\right\}$.

Explanation of Solution

Given information:

The equation is $3x^2-2x=-1$.

Formula used:

The solution of the equation $a{x}^2+bx+c=0$ is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

To get the equation $3x^2-2x=-1$ in standard form, add $1$ on both sides:

$3x^2-2x+1=-1+1$.

$\Rightarrow 3x^2-2x+1=0$

By using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

$\Rightarrow x=\frac{-(-2)\pm \sqrt{(-2)-4(3)(1)}}{2\left(3\right)}$

$\Rightarrow x=\frac{2\pm \sqrt{4-12}}{6}$

$\Rightarrow x=\frac{2\pm \sqrt{-8}}{6}$

$\Rightarrow x=\frac{2\pm 2\sqrt{2i}}{6}$.

Dividing numerator and denominator of the above expression $\frac{2\pm 2\sqrt{2i}}{6}$ by $2$,

$\Rightarrow x=\frac{1\pm \sqrt{2i}}{3}=\left\{\frac{1-\sqrt{2i}}{3},\frac{1+\sqrt{2i}}{3}\right\}$

Therefore, the solution set of the equation $3x^2-2x=-1$ is $\left\{\frac{1-\sqrt{2i}}{3},\frac{1+\sqrt{2i}}{3}\right\}$.