Question

How do you solve $\sqrt{3x+10}=x$ and find any extraneous solutions? \[\]

Answer 1

Hint: We recall that the square root function only takes non-negative value to have $3x+10\ge 0$. We square both sides of the given equation to obtain a quadratic equation. We solve the quadratic equation by splitting the middle term method to find the roots. If any of the obtained roots do not satisfy then they are extraneous. \[\]

Complete answer:
We know that the square root function takes only non-negative values as inputs and returns non-negative as outputs. We are given the following equation to solve
\[\sqrt{3x+10}=x\]
Since the square root only takes non-negative values we have
\[\begin{align}
  & 3x+10\ge 0 \\
 & \Rightarrow 3x\ge -10 \\
 & \Rightarrow x\ge \dfrac{-10}{3} \\
\end{align}\]
Since square root function also returns also non-negative values we have
\[\begin{align}
  & \sqrt{3x+10}\ge 0 \\
 & \Rightarrow x\ge 0 \\
\end{align}\]
So the condition for non-extraneous roots is $x\ge 0\text{ and x}\ge \dfrac{-10}{3}\Rightarrow x\ge 0$. We square both sides of the given equation to have;
\[\begin{align}
  & {{\left( \sqrt{3x+10} \right)}^{2}}={{x}^{2}} \\
 & \Rightarrow 3x+10={{x}^{2}} \\
 & \Rightarrow {{x}^{2}}-3x-10=0 \\
\end{align}\]
We see that the above equation is a quadratic equation with the middle term $-3$. We use splitting the middle term method to find $p,q$ such that $p+q=-3$ and $pq=1\times \left( -10 \right)=-10$. We have from factorization$10=2\times 5$. We can take $p=2,q=-5$ to satisfy the conditions. So we have
\[\begin{align}
  & \Rightarrow {{x}^{2}}+\left( -5+2 \right)x-10=0 \\
 & \Rightarrow {{x}^{2}}-5x+2x-10=0 \\
\end{align}\]
We take $x$ common from the first two terms and $2$common from the last two terms to have
\[\Rightarrow x\left( x-5 \right)+2\left( x-5 \right)=0\]
We take $x-5$ common in the above step to have;
\[\begin{align}
  & \Rightarrow \left( x-5 \right)\left( x+2 \right)=0 \\
 & \Rightarrow x-5=0\text{ or }x+2=0 \\
 & \Rightarrow x=5\text{ or }x=-2 \\
\end{align}\]
So we have obtained two solutions $x=5,-2$ but we have a condition on $x$ that $x\ge 0$. So the solution or root $x=-2$ is an extraneous solution and only solution is $x=5$.\[\]

Note: We note that an extraneous root or solution is a root which does not satisfy the original equation. The general quadratic equation is given by $a{{x}^{2}}+bx+c=0\left( a\ne 0 \right)$. We solve it by factorization by splitting the middle term with $p,q$ such that $p+q=b,pq=c\times a$. We can alternatively solve using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.