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How do you solve $\sqrt{11x+3}-2x=0$ ?

seo-qna
Last updated date: 25th Feb 2024
Total views: 339k
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IVSAT 2024
Answer
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Hint: We are given an equation as $\sqrt{11x+3}-2x=0$ we have to solve this problem, we will first understand what does solution mean then we will look for the type of equation. we have according to the type we will solve and simplify. We will remove the square root of the equation then we will find the quadratic equation $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve for ‘x’, we will check then which value of ‘x’ will be best suited.

Complete step by step answer:
We have $\sqrt{11x+3}-2x=0$, we have to find the solution. Solutions are those values of ‘x’ for which the given equation is satisfied.
Now to find the solution, we will simplify our problem, we will remove the square root from it.
To do so, we will first add ‘2x’ on both side, so we get –
$\sqrt{11x+3}-2x+2x=2x$
So, \[\sqrt{11x+3}=2x\]
Now squaring both side, we get –
\[{{\left( \sqrt{11x+3} \right)}^{2}}={{\left( 2x \right)}^{2}}\]
As ${{\left( \sqrt{11x+3} \right)}^{2}}=11x+3$ and ${{\left( 2x \right)}^{2}}=4{{x}^{2}}$ , so we get –
$11x+3=4{{x}^{2}}$ .
Now it has power ‘2’, so it is a quadratic equation, now changing it into standard form given as $a{{x}^{2}}+bx+c=0$
We will get –
$4{{x}^{2}}-11x-3=0$
Now we will use the quadratic formula of the quadratic equation to solve for the value of ‘x’.
For $a{{x}^{2}}+bx+c=0$ , Quadratic formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
For $4{{x}^{2}}-11x-3=0$
We have $a=4,b=-11\text{ and }c=-3$ so, solution will be –
 $x=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\times 4\times \left( -3 \right)}}{2\left( 4 \right)}$
By simplifying, we get –
$x=\dfrac{+11\pm \sqrt{169}}{8}$
As $\sqrt{169}=13$ so –
$x=\dfrac{11\pm 13}{8}$
Hence, solution are –
$x=\dfrac{-2}{8}\text{ and }x=\dfrac{24}{8}$
By simplifying further, we get –
$x=-0.25\text{ and }x=3$
Now we will see which of the two satisfy the equation.
We put $x=-0.25$ in $\sqrt{11x+3}-2x=0$ .
We get –
$\sqrt{11\times -0.25+3}-2\left( -0.25 \right)=0$
$\sqrt{-2.75+3}+0.5=0$ (as $-2\left( -0.25 \right)=+0.5$ )
$\sqrt{0.25}+0.5=0$
By simplifying, we get –
$0.5+0.5=0\Rightarrow 1=0$ , not true.
So, $x=-0.25$ is not the solution.
Now we check x=3
So, putting x=3 in $\sqrt{11x+3}-2x=0$we get –
$\begin{align}
  & \sqrt{11\times 3+3}-2\times 3=0 \\
 & \sqrt{36}-6=0 \\
 & 6-6=0\text{ }\left[ as\text{ }\sqrt{36}=6 \right] \\
 & 0=0 \\
\end{align}$
This is true.
So, x=3 is the only solution.

Note: Remember that our initial equation was $\sqrt{11x+3}-2x=0$, its highest power is ‘1’ so it can have at most zero only so stopping solution at $x=-0.25$ and $x=3$ will be in correct as out of these ‘2’ only ‘1’ will be correct. So we have to check for the correct one.
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