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# Solve the following quadratic equation by completing the square method ${{x}^{2}}+11x+30=0$.

Last updated date: 24th Jul 2024
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Hint: Try to write the 11x in the form of 2ab. Rearrange them in LHS and RHS in order to get two perfect squares on both sides. Apply general algebra rules and extract the roots.

Given equation is ${{x}^{2}}+11x+30=0$.
Or, we can write ${{x}^{2}}+2(\dfrac{11}{2})x+30=0$.
$\Rightarrow$ ${{x}^{2}}+2(\dfrac{11}{2})x+{{(\dfrac{11}{2})}^{2}}+30-{{(\dfrac{11}{2})}^{2}}=0$ [Adding ${{\left( \dfrac{11}{2} \right)}^{2}}$ on both sides]

Now, we know that,${{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}$. Therefore, we can write the LHS in this form.

So, ${{(x+\dfrac{11}{2})}^{2}}=\dfrac{121}{4}-30=\dfrac{121-120}{4}=\dfrac{1}{4}$

Now, taking square roots on both sides, we get two equations as taking square roots gives both positive and negative values.
$x+\dfrac{11}{2}=\dfrac{1}{2}$ ………. (1) and $x+\dfrac{11}{2}=-\dfrac{1}{2}$………… (2)

From (1) we get $x=\dfrac{1}{2}-\dfrac{11}{2}=\dfrac{1-11}{2}=-\dfrac{10}{2}=-5$ and from (2) we get $x=-\dfrac{1}{2}-\dfrac{11}{2}=-(\dfrac{1+11}{2})=-\dfrac{12}{2}=-6$.

These are the two roots of the given quadratic equation, which can be verified by putting the values of the roots in place of x and satisfying the equation.
Therefore, the roots of the given equation are -5 and -6 respectively.

Note: A quadratic equation must have at most (real/imaginary) two roots (all not necessarily distinct). Therefore, while taking square roots we should consider both positive and negative values. We can derive the well-known quadratic law for finding roots from this completing the square method. This method is the most fundamental method.