Chapter 7.4, Problem 18WE

To determine

Solve each equation by squaring both sides of the equation and then by using a quadratic form. Then we have to check whether either method give an extraneous root

Answer to Problem 18WE

  $x=9,1$

Explanation of Solution

Given information :

  $x+3=4\sqrt{x}$

Firstly we take square on both sides, we get,

  ${\left(x+3\right)}^2=16x$

Here we may use the formula ${\left(a+b\right)}^2=a^2+b^2+2ab$

Now we get,

  $\begin{array}{l}{\left(x+3\right)}^2=16x\\ x^2+9+6x=16x\\ x^2-10x+9=0\end{array}$

Now we know that the roots of the equation $a{x}^2+bx+c$ is given by $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence we compare the equation and put values we get,

  $\begin{array}{l}x=\frac{10\pm \sqrt{{\left(-10\right)}^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}\\ x=\frac{10\pm \sqrt{64}}{2}\\ x=\frac{10\pm 8}{2}\\ x=9,1\end{array}$

Here both 9 and 1 does satisfy when checked in original equation.