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# How do you solve ${{x}^{2}}-8x+16=0$ using a quadratic equation formula ?

Last updated date: 19th Sep 2024
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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.