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# How do you solve the quadratic equation by completing the square: ${{x}^{2}}-8x+12=0$ ?

Last updated date: 18th Jun 2024
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Hint: We are given a equation as ${{x}^{2}}-8x+12=0$ , To find the solution, we will first know about the type of equation once we have the type of equation their we will work on ways to find the solution completing the square method is one in which we make a complete square and then solve further, so we use this to solve our problem.

We are given an equation as ${{x}^{2}}-8x+12=0$ .
We can see that this equation has the highest power as 2, so it is clearly a two degree polynomial, a quadratic polynomial we have to find the solution of it, it means we have to find those in which we will satisfy the given equation.
As the power is 2 so maximum it can have 2 solutions.
We have to use completing the square method to find the solution.
The steps to find the solution using completing the square method is –
Step 1: Divide all term by coefficient of ${{x}^{2}}$ as in our equation ${{x}^{2}}-8x+12=0$
Coefficient of ${{x}^{2}}$ is ‘1’, so if we divide all terms, we get an equation same as the original.
${{x}^{2}}-8x+12=0$
Step 2: Move the number (constant) to the side in our equation we have constant as $+12$ so we subtract 12 on both sides, we get –
${{x}^{2}}-8x+12-12=0-12$
So, ${{x}^{2}}-8x=-12$
Step 3: Complete the square on the left side of the equation and balance this by adding the same value on the right.
In our equation ${{x}^{2}}-2\times 4\times x=-12$
We add ${{4}^{2}}$ to complete the square, we add it to both sides to keep the equation balanced.
So, ${{x}^{2}}-2\times 4x+{{4}^{2}}=12+{{4}^{2}}$
As ${{4}^{4}}=16$ so, we get at the simplifying
${{\left( x-4 \right)}^{2}}=4$
Taking square root we get –
$\left( x-4 \right)=\pm \sqrt{4}$
As $\sqrt{4}=2$
So,
$x-4=\pm 2$
Adding ‘4’ both side we get –
$x=\pm 2+4$
So, solutions are –
$x=+2+4$ and $x=-2+4$
So, $x=6$ and $x=2$

Note: We can also find a solution to this by plotting ${{x}^{2}}-8x+12=0$ on the graph. The point where there graph cut the x-axis will be defining our solution at here right side is zero so here solution is also refer to as zero if the given polynomial so that is why when we find the solution it will lie on the a-axis only as they are zero of the equation and zero always lie on x-axis.