Answer

Verified

372.9k+ 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.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail