Chapter 7.4, Problem 20WE

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 20WE

  $x=1,\frac{9}{4}$

Explanation of Solution

Given information:

  $2t+3=5\sqrt{t}$

Firstly we take square on both sides, we get,

  ${\left(2t+3\right)}^2=25t$

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

Now we get,

  $\begin{array}{l}4t^2+9+12t=25t\\ 4t^2-13t+9=0\end{array}$

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,

  $\begin{array}{l}x=\frac{13\pm \sqrt{{\left(-13\right)}^2-4\left(4\right)\left(9\right)}}{2\left(4\right)}\\ x=\frac{13\pm \sqrt{25}}{8}\\ x=\frac{13\pm 5}{8}\\ x=1,\frac{9}{4}\end{array}$

Here both roots satisfy when checked in original equation.