Question

How do you solve $\sqrt{3x+4}=10$?

Answer 1

Hint: We first take the square of the equation to find the linear form. Without taking squares we can’t solve the root part. Then using the binary operations, we find the solution for x.

Complete step by step answer:
We have to solve $\sqrt{3x+4}=10$.
We cannot solve the equation until we eliminate the square root part of the variables.
Therefore, to eliminate the root part we will take the square value of the equation for both sides.
We find the square of the equation $\sqrt{3x+4}=10$. We get ${{\left( \sqrt{3x+4} \right)}^{2}}={{10}^{2}}$.
Now for indices We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt[2]{a}=\sqrt{a}\].
Then squaring we get \[{{\left( {{a}^{\dfrac{1}{2}}} \right)}^{2}}={{\left( \sqrt{a} \right)}^{2}}=a\].
Then we can also find that ${{\left( \sqrt{3x+4} \right)}^{2}}=3x+4$.
Squaring the right-hand side, we get ${{10}^{2}}=100$.
The squared value is
$\begin{align}
  & {{\left( \sqrt{3x+4} \right)}^{2}}={{10}^{2}} \\
 & \Rightarrow 3x+4=100 \\
\end{align}$
Therefore, we get a linear equation of x.
We first subtract 4 from both sides to get
$\begin{align}
  & 3x+4-4=100-4 \\
 & \Rightarrow 3x=96 \\
\end{align}$
Now we divide both sides with 3 to get
$\begin{align}
  & \dfrac{3x}{3}=\dfrac{96}{3} \\
 & \Rightarrow x=32 \\
\end{align}$

Therefore, the solution of the equation $\sqrt{3x+4}=10$ is $x=32$.

Note: We always need to cross-check the value of x in the equation of $\sqrt{3x+4}=10$. For the equation $\sqrt{3x+4}=10$, when we put $x=32$, we get
$\sqrt{3x+4}=\sqrt{3\times 32+4}=\sqrt{100}=10$.
Therefore, the solution is justified.