# Solve the following equation: $\sqrt{6-4x-{{x}^{2}}}=x+4$.

Answer

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**Hint:**For solving this question you should know about the general mathematics calculations of expressions. In this problem it is asked to find the value of $x$. And this can be solved by general addition and subtraction of any numbers of terms on both sides. As we can say any term is added or subtracted as a form of zero or by making a square of both.

**Complete step by step answer:**

According to our question it is asked of us to find the value of $x$ if $\sqrt{6-4x-{{x}^{2}}}=x+4$. Now as we know that if we want to solve any equation or any expression, then we can add any term as a form of zero. It means that we will be doing addition and subtraction of the same term with the same signs, if we are adding or subtracting that from both sides of an equation. And if we add in only one side of the equation, then we add and subtract the same term with one negative and one positive sign. Many equations are solved by making the square of both sides if they are given the form of one side under root and one side normal in any equation. And we will use this here. So, if,

$\sqrt{6-4x-{{x}^{2}}}=x+4$

Then make a square of both sides. So,

$\begin{align}

& \Rightarrow 6-4x-{{x}^{2}}={{\left( x+4 \right)}^{2}} \\

& \Rightarrow 6-4x-{{x}^{2}}={{x}^{2}}+8x+16 \\

& \Rightarrow 2{{x}^{2}}-12x+10=0 \\

\end{align}$

Now, if we divide it by 2, then,

$\Rightarrow {{x}^{2}}-6x+5=0$

If we make the factored form of this, then,

$\begin{align}

& \Rightarrow \left( x-5 \right)\left( x-1 \right)=0 \\

& \Rightarrow x=5,1 \\

\end{align}$

So, the values of $x$ are 5 and 1.

**Note:**While solving this type of questions we have to always add or subtract the term which will make it easy for the question. And if the variables are of higher order, then too we will make the higher ordered term to add or subtract in the equation.

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