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Last updated date: 05th Mar 2024
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IVSAT 2024
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Hint: We first take the square on both sides of the given equation $\sqrt{x}=x-6$ to form the quadratic form. We use both grouping methods and vanishing methods to solve the problem. We take common terms out to form the multiplied forms. Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number. In the case of vanishing method, we use the value of $x$ which gives the polynomial value 0.

Complete step by step answer:
We first take the square on both sides of the given equation $\sqrt{x}=x-6$ to form the quadratic form. We get
  & {{\left( \sqrt{x} \right)}^{2}}={{\left( x-6 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}-13x+36=0 \\
We apply the middle-term factoring or grouping to factorise the polynomial.
In the case of \[{{x}^{2}}-13x+36\], we break the middle term $-13x$ into two parts of $-9x$ and $-4x$.
So, ${{x}^{2}}-13x+36={{x}^{2}}-9x-4x+36$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variable, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases give $36{{x}^{2}}$. The grouping will be done for ${{x}^{2}}-9x$ and $-4x+36$.
We try to take the common numbers out.
For ${{x}^{2}}-9x$, we take $x$ and get $x\left( x-9 \right)$.
For $-4x+36$, we take $-4$ and get $-4\left( x-9 \right)$.
The equation becomes ${{x}^{2}}-13x+36={{x}^{2}}-9x-4x+36=x\left( x-9 \right)-4\left( x-9 \right)$.
Both the terms have $\left( x-9 \right)$ in common. We take that term again and get
  & {{x}^{2}}-13x+36 \\
 & =x\left( x-9 \right)-4\left( x-9 \right) \\
 & =\left( x-9 \right)\left( x-4 \right) \\
Therefore, $\left( x-9 \right)\left( x-4 \right)=0$ has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
So, values of x are $x=4,9$.
Now we put those values in the equation $\sqrt{x}=x-6$ to verify
So, putting 4 in $\sqrt{x}=x-6$, we get
  & \sqrt{4}=4-6 \\
 & \Rightarrow 2=-2 \\
This is not possible. The root value of 4 is wrong.
putting 9 in $\sqrt{x}=x-6$, we get
  & \sqrt{9}=9-6 \\
 & \Rightarrow 3=3 \\
This is possible. The root value of 9 is right.

Therefore, the solution for $\sqrt{x}=x-6$ is $x=9$.

Note: We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+4x-32=0$. The values of a, b, c is $1,4,-32$ respectively.
We put the values and get x as $x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 1\times \left( -32 \right)}}{2\times 1}=\dfrac{-4\pm \sqrt{144}}{2}=\dfrac{-4\pm 12}{2}=4,-8$.
We got an extra solution for squaring the equation at the start.
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