Question

Solve the following equations for $x$:

$\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}$

Answer 1

Step 1:

Reduce the given equation to standard quadratic equation form:

Taking square on both sides of the given equation,

$\begin{array}{rcl}{\left(\sqrt{x+5}+\sqrt{x+21}\right)}^2& =& {\left(\sqrt{6x+40}\right)}^2\\ & \Rightarrow & \left(x+5\right)+\left(x+21\right)+2\sqrt{x+5}\times \sqrt{x+21}=6x+40\\ & \Rightarrow & 2\sqrt{x+5}\times \sqrt{x+21}=6x-x-x+40-5-21\\ & \Rightarrow & 2\sqrt{x+5}\times \sqrt{x+21}=4x+14\\ & \Rightarrow & \sqrt{\left(x+5\right)\left(x+21\right)}=2x+7\left(\mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}2\right)\\ & \Rightarrow & \left(x+5\right)\left(x+21\right)={\left(2x+7\right)}^2\left(\mathrm{Square}\mathrm{on}\mathrm{both}\mathrm{sides}\right)\\ & \Rightarrow & x^2+21x+5x+105=4x^2+28x+49\\ & \Rightarrow & 3x^2+2x+49-105=0\\ & \Rightarrow & 3x^2+2x-56=0\end{array}$

Step 2:

Determine the value of $x$:

Apply factorization method to solve the obtained quadratic equation.

$$\begin{gathered} 3x^2+2x-56=0 \\ \begin{array}{rcl}& \Rightarrow & 3x^2+14x-12x-56=0\\ & \Rightarrow & x\left(3x+14\right)-4\left(3x+14\right)=0\\ & \Rightarrow & \left(x-4\right)\left(3x+14\right)=0\\ & \Rightarrow & x-4=0\mathrm{or}3x+14=0\\ & \Rightarrow & x=4\mathrm{or}x=-\frac{14}{3}\end{array} \end{gathered}$$

Verify the obtained values of $x$.

When $x=-\frac{14}{3}$,

$\begin{array}{rcl}\sqrt{x+5}+\sqrt{x+21}& =& \sqrt{\left(-\frac{14}{3}\right)+5}+\sqrt{\left(-\frac{14}{3}\right)+21}\\ & =& \sqrt{1}+\sqrt{49}\\ & =& 8\left[\mathrm{Principal}\mathrm{root}\mathrm{symbol}\mathrm{give}\mathrm{positive}\mathrm{values}\mathrm{only}\right]\\ \sqrt{6x+40}& =& \sqrt{6\times \left(-\frac{14}{3}\right)+40}\\ & =& \sqrt{-28+40}\\ & =& \sqrt{16}\\ & =& 4\left[\mathrm{Principal}\mathrm{root}\mathrm{symbol}\mathrm{give}\mathrm{positive}\mathrm{values}\mathrm{only}\right]\\ \therefore \sqrt{x+5}+\sqrt{x+21}& \ne & \sqrt{6x+40}\end{array}$

Hence, $x=-\frac{14}{3}$ is not a solution to the given equation.

When $x=4$,

$\begin{array}{rcl}\sqrt{x+5}+\sqrt{x+21}& =& \sqrt{4+5}+\sqrt{4+21}\\ & =& \sqrt{9}+\sqrt{25}\\ & =& 8\left[\mathrm{Principal}\mathrm{root}\mathrm{symbol}\mathrm{give}\mathrm{positive}\mathrm{values}\mathrm{only}\right]\\ \sqrt{6x+40}& =& \sqrt{6\times \left(4\right)+40}\\ & =& \sqrt{24+40}\\ & =& \sqrt{64}\\ & =& 8\left[\mathrm{Principal}\mathrm{root}\mathrm{symbol}\mathrm{give}\mathrm{positive}\mathrm{values}\mathrm{only}\right]\\ \therefore \sqrt{x+5}+\sqrt{x+21}& =& \sqrt{6x+40}\end{array}$

Hence, the required solution is $x=4$.