Answer

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**Hint:**In order to solve this question, we will solve the equation as a quadratic equation and we will simplify the equation by finding the roots and by applying quadratic equation formula that is, $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

**Complete step by step solution:**

We have the given equation:

\[{{x}^{2}}-8x+16=0\]

Now for a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following quadratic equation formula: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Therefore from the above equation and on comparing, we get $a=1,b=-8,c=16$

Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 1\times \left( 16 \right)}}{2\times 1}$

Now we will first simplify, root part in our equation:

Now, we will apply the exponent rule: ${{\left( -a \right)}^{n}}={{a}^{n}}$, if n is even.

Therefore, we get:

$\Rightarrow {{\left( -8 \right)}^{2}}={{\left( 8 \right)}^{2}}$

So, we get our equation as:

$\Rightarrow \sqrt{{{\left( 8 \right)}^{2}}-4\times 1\times 16}$

On simplifying:

$\Rightarrow \sqrt{64-64}$

$\Rightarrow \sqrt{0}$

Now coming to our equation and substituting what we equate, we get:

\[\] $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{8\pm \sqrt{0}}{2}$

Now, on simplifying above equation, we get:

$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{8}{2}$

Now as we can see that there is a common multiple of $2$, so we will use factorization method.

Therefore, on factorization that is on dividing and equating in the equation, we get:

$\left( {{x}_{1}},{{x}_{2}} \right)=4$

Now on equating the above equation we get the value of x, i.e.:

${{x}_{1}}=4$ and ${{x}_{2}}=4$

Therefore,

$x=4$ is the solution of the given quadratic equation.

**Note:**

There are three forms of quadratic equation, Standard form:$y=a{{x}^{2}}+bx+c$ where $a,b,c$ are just numbers. Factored form:\[~y\text{ }=\text{ }\left( ax\text{ }+\text{ }c \right)\left( bx\text{ }+\text{ }d \right)\] again the $a,b,c,d$ are just numbers. Vertex form: \[y\text{ }=\text{ }a{{\left( x\text{ }+\text{ }b \right)}^{2}}\text{ }+\text{ }c\] again the $a,b,c$ are just numbers. We have to identify by looking at the equation which form we have to use, then only we can solve the equation.

As we have a standard form of quadratic equation, we can verify the answer by substituting the answer in the original equation.

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