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# How do you find extraneous solutions from radical equations?

Last updated date: 13th Jun 2024
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Hint: An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation. We get these types of solutions sometimes when we solve radical equations. Radical equations are equations in which the variable is under a radical. To find the solution of a radical equation, we need to take the radical expression to one side of the equations. If there is more than one radical expression, take one at a time.

Let’s take an example of radical expression $x+1=\sqrt{7x+15}$. We need to solve this equation. Squaring both sides of the above equation, we get
$\Rightarrow {{\left( x+1 \right)}^{2}}={{\left( \sqrt{7x+15} \right)}^{2}}$
simplifying the above equation, we get
\begin{align} & \Rightarrow {{x}^{2}}+2x+1=7x+15 \\ & \Rightarrow {{x}^{2}}-5x-14=0 \\ \end{align}
We can find the roots of the above quadratic equation, by using the formula method, as follows
\begin{align} & \Rightarrow x=\dfrac{-(-5)\pm \sqrt{{{\left( -5 \right)}^{2}}-4(1)(-14)}}{2(1)} \\ & \Rightarrow x=\dfrac{5\pm \sqrt{81}}{2} \\ & \Rightarrow x=\dfrac{5\pm 9}{2} \\ \end{align}
$\Rightarrow x=\dfrac{5+9}{2}=\dfrac{14}{2}$ or $x=\dfrac{5-9}{2}=\dfrac{-4}{2}$
$\therefore x=7$ or $x=-2$

But if we substitute $x=-2$ in the equation, we get
\begin{align} & -2+1=\sqrt{7(-2)+15} \\ & \Rightarrow -1=1 \\ \end{align}
Which is not correct. Hence, $x=-2$ is not a solution of the given radical equation. Thus $x=-2$ is an extraneous solution for the given equation.

Note: We should know when an extraneous solution occurs. Extraneous solutions of an equation are solutions that occur when a radical expression that has an even index, such as 2, is raised to its power to find the solution of an equation.
In the above example, as the radical power has an even index, we get $x=-2$ as an extraneous solution of the equation.