Answer
397.5k+ views
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.
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)