How do you solve the expression $\sqrt{x+8}=6$ ?
Answer
565.8k+ views
Hint: We are given a equation as $\sqrt{x+8}=6$ , we have to solve this problem, we will first understand solution, what are mean by them, then we will try to look for those x which satisfies the equation. We will first simplify the expression by using algebraic tools, We can also graph the equation to find the solution.
Another way of approach will be hit and trial method; we put a value of ‘x’ which may satisfy the solution.
Complete step by step solution:
We are given $\sqrt{x+8}=6$ , we can see that the variable is under the root $\left( \sqrt{{}} \right)$ and if we simplify it, it can have maximum power as 1, so this equation $\sqrt{x+8}=6$ can have maximum of solution.
Solutions are referred to those ‘x’ which are then selected in the equation then they satisfy the equation.
Now to find those ‘x’,
We can use algebraic tools to find the solution.
As we have $\sqrt{x+8}=6$ .
By squaring on both sides, we get –
$\Rightarrow {{\left( \sqrt{x+8} \right)}^{2}}={{6}^{2}}$
By simplifying, we get –
$\Rightarrow x+8=36$ (as ${{6}^{2}}=36$ )
By subtracting 8 on both sides, we get –
$\Rightarrow x+8-8=36-8$
So, we get –
$x=28$ .
Hence we get $x=28$ in the solution of $\sqrt{x+8}=6$ .
Note: We can also solve this using another method.
We will go by hit and trial method.
We have $\sqrt{x+8}=6$ .
As we know square roots work not on negative value so we move along positive.
Now let $x=1$
We put $x=1$ in $\sqrt{x+8}=6$
We get –
$\sqrt{1+8}=6$
By simplifying, we get –
$\sqrt{9}=3$
As $\sqrt{9}=3$ so $3=6$
Which is not true.
Now, we put $x=8$ so we get –
$\begin{align}
& \Rightarrow \sqrt{8+8}=6 \\
& \sqrt{16}=6 \\
\end{align}$
As $\sqrt{16}=4$ so
$4=6$
Not true.
So, $x=8$ is not a solution.
Let we put $x=17$ , we get –
$\begin{align}
& \Rightarrow \sqrt{17+8}=6 \\
& \sqrt{25}=6 \\
\end{align}$
As $\sqrt{25}=5$ so,
$5=6$
Not true.
We are going correctly as the difference is decreasing.
Now, we let $x=28$ in $\sqrt{x+8}=6$ , we get –
$\begin{align}
& \Rightarrow \sqrt{28+8}=6 \\
& \sqrt{36}=6 \\
\end{align}$
$6=6$ (as $\sqrt{36}=6$ )
So, we get –
$x=28$ is the correct solution.
So, we get that the solution for $\sqrt{x+8}=6$ is $x=28$ .
Another way of approach will be hit and trial method; we put a value of ‘x’ which may satisfy the solution.
Complete step by step solution:
We are given $\sqrt{x+8}=6$ , we can see that the variable is under the root $\left( \sqrt{{}} \right)$ and if we simplify it, it can have maximum power as 1, so this equation $\sqrt{x+8}=6$ can have maximum of solution.
Solutions are referred to those ‘x’ which are then selected in the equation then they satisfy the equation.
Now to find those ‘x’,
We can use algebraic tools to find the solution.
As we have $\sqrt{x+8}=6$ .
By squaring on both sides, we get –
$\Rightarrow {{\left( \sqrt{x+8} \right)}^{2}}={{6}^{2}}$
By simplifying, we get –
$\Rightarrow x+8=36$ (as ${{6}^{2}}=36$ )
By subtracting 8 on both sides, we get –
$\Rightarrow x+8-8=36-8$
So, we get –
$x=28$ .
Hence we get $x=28$ in the solution of $\sqrt{x+8}=6$ .
Note: We can also solve this using another method.
We will go by hit and trial method.
We have $\sqrt{x+8}=6$ .
As we know square roots work not on negative value so we move along positive.
Now let $x=1$
We put $x=1$ in $\sqrt{x+8}=6$
We get –
$\sqrt{1+8}=6$
By simplifying, we get –
$\sqrt{9}=3$
As $\sqrt{9}=3$ so $3=6$
Which is not true.
Now, we put $x=8$ so we get –
$\begin{align}
& \Rightarrow \sqrt{8+8}=6 \\
& \sqrt{16}=6 \\
\end{align}$
As $\sqrt{16}=4$ so
$4=6$
Not true.
So, $x=8$ is not a solution.
Let we put $x=17$ , we get –
$\begin{align}
& \Rightarrow \sqrt{17+8}=6 \\
& \sqrt{25}=6 \\
\end{align}$
As $\sqrt{25}=5$ so,
$5=6$
Not true.
We are going correctly as the difference is decreasing.
Now, we let $x=28$ in $\sqrt{x+8}=6$ , we get –
$\begin{align}
& \Rightarrow \sqrt{28+8}=6 \\
& \sqrt{36}=6 \\
\end{align}$
$6=6$ (as $\sqrt{36}=6$ )
So, we get –
$x=28$ is the correct solution.
So, we get that the solution for $\sqrt{x+8}=6$ is $x=28$ .
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