How do you solve $\sqrt{2x+10}-6=2$
Answer
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Hint: Now we are given with the equation $\sqrt{2x+10}-6=2$ . To solve this equation we will first take 6 to RHS and change signs accordingly. Now we will square the whole equation. We know that ${{\left( \sqrt{x} \right)}^{2}}=x$ using this we will get one linear equation in x. Now we will again rearrange the whole equation and then divide the whole equation by 2. Hence we will get the value of x which is nothing but the solution of the given equation.
Complete step-by-step solution:
Now we are given with the equation $\sqrt{2x+10}-6=2$
Now let us take – 6 on RHS. Since we are taking the term other side hence the sign of the term will change which means – 6 will become + 6.
Hence we will get $\sqrt{2x+10}=2+6$
Hence we get, $\sqrt{2x+10}=8$
Now let us take a square on both sides of the above equation.
$\Rightarrow {{\left( \sqrt{2x+10} \right)}^{2}}={{8}^{2}}$
Now we know that ${{\left( \sqrt{x} \right)}^{2}}=x$ and ${{8}^{2}}=64$ .
Hence we have $2x+10=64$
Now let us take 10 on RHS. Now since we are shifting the term to another side we will change its sign. Hence, we get
$\Rightarrow 2x=64-10$
Doing subtraction on RHS we get,
$\Rightarrow 2x=54$
Now dividing the whole equation by 2 we get,
$\begin{align}
& \Rightarrow \dfrac{2x}{2}=\dfrac{54}{2} \\
& \Rightarrow x=27 \\
\end{align}$
Hence we get the value of x = 27.
Hence the solution of the given equation is x = 27.
Note: Since we have a square root involved in the equation we know we have to square the equation. Now if we use square in the first step itself then we will have to use the formula${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . We can solve the equation by this method too but this may complicate the solution and hence we first separate the square root on one side.
Complete step-by-step solution:
Now we are given with the equation $\sqrt{2x+10}-6=2$
Now let us take – 6 on RHS. Since we are taking the term other side hence the sign of the term will change which means – 6 will become + 6.
Hence we will get $\sqrt{2x+10}=2+6$
Hence we get, $\sqrt{2x+10}=8$
Now let us take a square on both sides of the above equation.
$\Rightarrow {{\left( \sqrt{2x+10} \right)}^{2}}={{8}^{2}}$
Now we know that ${{\left( \sqrt{x} \right)}^{2}}=x$ and ${{8}^{2}}=64$ .
Hence we have $2x+10=64$
Now let us take 10 on RHS. Now since we are shifting the term to another side we will change its sign. Hence, we get
$\Rightarrow 2x=64-10$
Doing subtraction on RHS we get,
$\Rightarrow 2x=54$
Now dividing the whole equation by 2 we get,
$\begin{align}
& \Rightarrow \dfrac{2x}{2}=\dfrac{54}{2} \\
& \Rightarrow x=27 \\
\end{align}$
Hence we get the value of x = 27.
Hence the solution of the given equation is x = 27.
Note: Since we have a square root involved in the equation we know we have to square the equation. Now if we use square in the first step itself then we will have to use the formula${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . We can solve the equation by this method too but this may complicate the solution and hence we first separate the square root on one side.
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