Chapter 5.6, Problem 11Q

Solution

  

To calculate: Solve $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$ equation and check extraneous solution.

Solution of $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$ is $x=\left(2+\sqrt{7}\right)$ and $x=\left(2-\sqrt{7}\right)$.

Given information: Given $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$ . Where x is variable.

Calculation: Solution of:

  $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$

  $\frac{5+x+1}{x\left(x+2\right)}=\frac{2x+9}{x+2}$

Cut $\left(x+2\right)$ on both side in above equation.

  $\frac{5+x+1}{x}=\frac{2x+9}{1}$

  $\frac{6+x}{x}=\frac{2x+9}{1}$

By using cross multiplication solve this equation

Now,

  $\left(x+6\right)=x\times \left(2x+9\right)$

  $\left(x+6\right)=2x^2+9x$

  $2x^2+9x-x-6=0$

  $2x^2+8x-6=0$

  $2\left(x^2+4x-3\right)=0$

  $\left(x^2+4x-3\right)=0$

Now solve this quadratic equation. By using biquadratic formula.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Compare this equation $\left(x^2+4x-3\right)=0$ with $\left(a{x}^2+bx+c\right)=0$

Now a=1 b=4 c=-3

Put this value in formula

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

  $x=\frac{4\pm \sqrt{\left(16\right)+12}}{2}$

  $x=\frac{4\pm \sqrt{28}}{2}$

  $x=2\left(\frac{2\pm \sqrt{7}}{2}\right)$

  $x=\left(2\pm \sqrt{7}\right)$

  $x=\left(2+\sqrt{7}\right)$ And $x=\left(2-\sqrt{7}\right)$

Thus Solution of $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$ is $x=\left(2+\sqrt{7}\right)$ and $x=\left(2-\sqrt{7}\right)$

Now check this solution is extraneous or not.

Put this value in original equation $\frac{5}{x}+\frac{x+1}{x+2}=\frac{2x+9}{x+2}$

Hence $x=\left(2+\sqrt{7}\right)$ and $x=\left(2-\sqrt{7}\right)$ is not extraneous solution.