Question

How do you solve ${{x}^{2}}-12x+5=0$ by completing the square?

Answer 1

Hint: To solve a quadratic equation $a{{x}^{2}}+bx+c$ by completing square method we have to change the value of constant term such that the expression will be a square term. Then we can solve the equation by taking the square root in LHS and RHS.

Complete step by step answer:
Let’s assume we have a quadratic equation $a{{x}^{2}}+bx+c=0$ to convert it into square term lets multiply the term with a so that the coefficient of ${{x}^{2}}$ will became a square number
${{a}^{2}}{{x}^{2}}+abx+ac=0$
$\Rightarrow {{a}^{2}}{{x}^{2}}+2\times \dfrac{1}{2}b\times ax+ac=0$
We know that ${{\left( m+n \right)}^{2}}={{m}^{2}}+2mn+{{n}^{2}}$
In our equation ${{m}^{2}}={{a}^{2}}{{x}^{2}}$ so $m=ax$ and from the middle term we can tell that $n=\dfrac{b}{2}$ .
So the constant term that is in our case $ac$ should me change to ${{\left( \dfrac{b}{2} \right)}^{2}}$ adding and subtracting
$\dfrac{{{b}^{2}}}{4}$ in LHS
$\Rightarrow {{a}^{2}}{{x}^{2}}+2\times \dfrac{1}{2}b\times ax+ac+\dfrac{{{b}^{2}}}{4}-\dfrac{{{b}^{2}}}{4}=0$
$\Rightarrow {{a}^{2}}{{x}^{2}}+2\times \dfrac{1}{2}b\times ax+\dfrac{{{b}^{4}}}{4}=\dfrac{{{b}^{2}}}{4}-ac$
Now we can clearly see LHS is square term with $m=ax$ and $n=\dfrac{b}{2}$ so further solving
$\Rightarrow {{\left( ax+\dfrac{b}{2} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4}$
Taking square root both sides
$\Rightarrow ax+\dfrac{b}{2}=\dfrac{\pm \sqrt{{{b}^{2}}-4ac}}{2}$
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So in our case the quadratic equation is
${{x}^{2}}-12x+5=0$
$a=1,b=-12,c=5$
${{x}^{2}}-2\times 6\times x+5=0$
We can add 36 in both LHS and RHS
${{x}^{2}}-2\times 6\times x+36+5=36$
$\Rightarrow {{x}^{2}}-2\times 6\times x+36=31$
Now the LHS is a square term
${{\left( x-6 \right)}^{2}}=31$
$\Rightarrow x-6=\pm \sqrt{31}$
$\Rightarrow x=6\pm \sqrt{31}$
If we observe we can see we first multiply the whole equation with a to convert the coefficient of the ${{x}^{2}}$ into square number in that we could have divided the whole term by a so that $a{{x}^{2}}$ became ${{x}^{2}}$ then repeat the same procedure after that we will get the same result.

Note: We can see that the formula for finding the roots of a quadratic equation is derived from completing the square method. If we try to solve a quadratic equation having complex roots by complete square method, the LHS will be square term but the RHS will be a negative number that results in the roots to be complex.