Chapter 8.6, Problem 20PPS

To determine

To solve : the equation

Answer to Problem 20PPS

The solution of the equation is $x=-5\pm \sqrt{7}$

Explanation of Solution

Given information :

The equation is

  $\frac{x}{2x-1}+\frac{3}{x+4}=\frac{-21}{2x^2+7x-4}$

Formula used : Sridhar Acharya $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation :

The LCD for the term is $\left(2x-1\right)\left(x+4\right)$ .

  $\begin{array}{l}\frac{x}{2x-1}+\frac{3}{x+4}=\frac{-21}{2x^2+7x-4}\\ \left(2x-1\right)\left(x+4\right)\left(\frac{x}{2x-1}\right)+\left(2x-1\right)\left(x+4\right)\left(\frac{3}{x+4}\right)=\left(2x-1\right)\left(x+4\right)\left(\frac{-21}{2x^2+7x-4}\right)\\ x\left(x+4\right)+3\left(2x-1\right)=\left(2x^2+8x-x-4\right)\left(\frac{-21}{2x^2+7x-4}\right)\\ x\left(x+4\right)+3\left(2x-1\right)=\left(2x^2+7x-4\right)\left(\frac{-21}{2x^2+7x-4}\right)\\ x^2+4x+6x-3=-21\\ x^2+10x-3+21=0\\ x^2+10x+81=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-10\pm \sqrt{{\left(10\right)}^2-4\cdot 1\cdot 18}}{2\cdot 1}\\ x=\frac{-10\pm \sqrt{100-72}}{2}\\ x=\frac{-10\pm \sqrt{28}}{2}\\ x=\frac{-10\pm 2\sqrt{7}}{2}\\ x=-5\pm \sqrt{7}\end{array}$

Check $\frac{x}{2x-1}+\frac{3}{x+4}=\frac{-21}{2x^2+7x-4}$

Take positive sing

  $\begin{array}{l}\frac{-5+\sqrt{7}}{2\left(-5+\sqrt{7}\right)-1}+\frac{3}{\left(-5+\sqrt{7}\right)+4}=\frac{-21}{2{\left(-5+\sqrt{7}\right)}^2+7\left(-5+\sqrt{7}\right)-4}\\ \frac{-5+\sqrt{7}}{-10+2\sqrt{7}-1}+\frac{3}{-1+\sqrt{7}}=\frac{-21}{2\{{\left(-5\right)}^2+2\left(-5\right)\left(\sqrt{7}\right)+{\left(\sqrt{7}\right)}^2\}-35+7\sqrt{7}-4}\\ \frac{-5+\sqrt{7}}{-11+2\sqrt{7}}+\frac{3}{-1+\sqrt{7}}=\frac{-21}{2\left(25-10\sqrt{7}+7\right)-39+7\sqrt{7}}\\ \frac{\left(-5+\sqrt{7}\right)\left(-1+\sqrt{7}\right)+3\left(-11+2\sqrt{7}\right)}{\left(-11+2\sqrt{7}\right)\left(-1+\sqrt{7}\right)}=\frac{-21}{2\left(32-10\sqrt{7}\right)-39+7\sqrt{7}}\\ \frac{5-\sqrt{7}-5\sqrt{7}+7-33+6\sqrt{7}}{11-2\sqrt{7}-11\sqrt{7}+2\sqrt{7}\left(\sqrt{7}\right)}=\frac{-21}{64-20\sqrt{7}-39+7\sqrt{7}}\\ \frac{-6\sqrt{7}-21+6\sqrt{7}}{11-13\sqrt{7}+14}=\frac{-21}{25-13\sqrt{7}}\\ \frac{-21}{25-13\sqrt{7}}=\frac{-21}{25-13\sqrt{7}}\end{array}$

Take negative sing.

  $\begin{array}{l}\frac{-5-\sqrt{7}}{2\left(-5-\sqrt{7}\right)-1}+\frac{3}{\left(-5-\sqrt{7}\right)+4}=\frac{-21}{2{\left(-5-\sqrt{7}\right)}^2+7\left(-5-\sqrt{7}\right)-4}\\ \frac{-5+\sqrt{7}}{-10-2\sqrt{7}-1}+\frac{3}{-1-\sqrt{7}}=\frac{-21}{2\{{\left(-5\right)}^2-2\left(-5\right)\left(\sqrt{7}\right)+{\left(\sqrt{7}\right)}^2\}-35-7\sqrt{7}-4}\\ \frac{-5-\sqrt{7}}{-11-2\sqrt{7}}+\frac{3}{-1-\sqrt{7}}=\frac{-21}{2\left(25+10\sqrt{7}+7\right)-39-7\sqrt{7}}\\ \frac{\left(-5-\sqrt{7}\right)\left(-1-\sqrt{7}\right)+3\left(-11-2\sqrt{7}\right)}{\left(-11-2\sqrt{7}\right)\left(-1-\sqrt{7}\right)}=\frac{-21}{2\left(32+10\sqrt{7}\right)-39-7\sqrt{7}}\\ \frac{5+5\sqrt{7}+\sqrt{7}+\left(\sqrt{7}\cdot \sqrt{7}\right)-33-6\sqrt{7}}{11+2\sqrt{7}+11\sqrt{7}+2\sqrt{7}\left(\sqrt{7}\right)}=\frac{-21}{64+20\sqrt{7}-39-7\sqrt{7}}\\ \frac{5+6\sqrt{7}+7-33-6\sqrt{7}}{11+13\sqrt{7}+14}=\frac{-21}{25+13\sqrt{7}}\\ \frac{-21}{25+13\sqrt{7}}=\frac{-21}{25+13\sqrt{7}}\end{array}$

Hence, the solution of the equation is $x=-5\pm \sqrt{7}$ .