Question

How do you solve $\sqrt{9x+10}=x$ and find any extraneous solutions?

Answer 1

To solve these questions with a radical sign over the terms on one side of the equation, first square both the sides and then solve the quadratic expression by using any of the methods and solve for the variable.

Complete step by step solution:
Given:
The equation $\sqrt{9x+10}=x$ .
Now, for the first step square both the sides of the given equation to get,
$\Rightarrow {{\left( \sqrt{9x+10} \right)}^{2}}={{\left( x \right)}^{2}}$
Rewriting the expression, we get,
$\Rightarrow {{x}^{2}}-9x-10=0$
Now, solving the above expression using splitting the middle term we get,
$\Rightarrow {{x}^{2}}-10x+x-10=0$
Since, the product of the first two terms was $-10$ and the middle term can be written as the sum of $-10x$ and $x$ .
Now, taking out common from the terms we get,
$\Rightarrow x\left( x-10 \right)+1\left( x-10 \right)=0$
Further on taking out the common expressions and simplifying it, we get,
$\Rightarrow \left( x+1 \right)\left( x-10 \right)=0$
From the above equation, we can get the value of the variable $x$ as,
$\Rightarrow x=-1,10$
So, the possible solutions can be given as,
$x=-1$ And $x=10$
Now, to find the extraneous points, substitute the values back in the original equation,
Case 1: when $x=-1$
Equation: $\sqrt{9x+10}=x$
Left-hand side
$\Rightarrow \sqrt{\left( 9\times -1 \right)+10}$
$\Rightarrow \sqrt{-9+10}$
$\Rightarrow \sqrt{1}=1$
And Right-hand side
$\Rightarrow x=-1$
Since, Right-hand side is not equal to the left-hand side of the equation, that is, $L.H.S\ne R.H.S$, $x=-1$ this is an extraneous solution.
Case 2: when $x=10$ ,
Equation: $\sqrt{9x+10}=x$
Left-hand side
$\Rightarrow \sqrt{9\left( 10 \right)+10}$
$\Rightarrow \sqrt{90+10}$
$\Rightarrow \sqrt{100}=10$
And Right-hand side
$\Rightarrow x=10$
Since, Right-Hand side is equal to the left-hand side of the equation, this is not an extraneous solution.
Therefore, after solving the equation $\sqrt{9x+10}=x$ the values of the variable are $x=-1$ and $x=10$ .
And after substituting the values back in the equation, we find that $x=-1$ is an extraneous solution.

Note: Remember that while substituting the values of the variable back in the equation, take the original equation, that is, the equation given in the question. An extraneous solution can be defined as that value of the variable which when substituted back in the equation does not make the left and the right hand side of the equation equal.