Chapter 12.5, Problem 27WE

To determine

Solve the quadratic equation.

Answer to Problem 27WE

  $x\in \left\{\frac{2}{11},-\frac{8}{5}\right\}$

Explanation of Solution

Given:

The equation is :

  ${\left(3x-5\right)}^2={\left(8x+3\right)}^2$

Calculation:

The given equation ${\left(3x-5\right)}^2={\left(8x+3\right)}^2$ .

So,

  $\begin{array}{l}{\left(3x-5\right)}^2={\left(8x+3\right)}^2\\ \Rightarrow 9x^2+25-30x=64x^2+9+48x\\ \Rightarrow 55x^2+78x-16=0\\ \Rightarrow 55x^2+88x-10x-16=\mathrm{0.....................}[\text{factor the middle term]}\\ \Rightarrow 11x(5x+8)-2(5x+8)=0\\ \Rightarrow (11x-2)(5x+8)=0\\ \Rightarrow 11x-2=0\text{ or }5x+8=\mathrm{0.................}[\text{zero product property]}\\ \Rightarrow x=\frac{2}{11}\text{ or }x=-\frac{8}{5}\end{array}$

Hence , $x\in \left\{\frac{2}{11},-\frac{8}{5}\right\}$