Question

Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
${x^2} - 4ax + 4{a^2} - {b^2} = 0$

Answer 1

Hint:Completing the square is a method to solve quadratic Equations by changing the form of the equation, so that the left side is a perfect square trinomial. In this method firstly divide the whole equation with coefficient of ${x^2}$, then half the coefficient of $x$ term to both sides. Factor the left side as a perfect square trinomial and take the square root of both sides and solve for $x$.

Complete step-by-step answer:
${x^2} - 4ax + 4{a^2} - {b^2} = 0$
Coefficient of ${x^2} = 1$
$\Rightarrow {x^2} - 4ax + 4{a^2} - {b^2} = 0$
Adding and subtracting $4{a^2}$ in both L.H.S and R.H.S
$\left[ {{x^2} - 2 \times x\left( {2a} \right) + {{\left( {2a} \right)}^2}} \right] + 4{a^2} - {b^2} = {\left( {2a} \right)^2} \\\$
We know the identity $(a-b)^2=a^2+b^2-2ab$
Here in this , where $a=x$ and $b=-2a$ converting into this form,we get
${\left( {x - 2a} \right)^2} = 4{a^2} - 4{a^2} + {b^2} \\\ {\left( {x - 2a} \right)^2} = {b^2} \\\$
Taking square root of both sides
$x - 2a = \pm b \\\ x = 2a \pm b \\\ \Rightarrow x = 2a + b{\text{ }}or{\text{ }}x = 2a - b \\\$

Note- This problem can also be solved by the direct use of the formula for the root of the quadratic equation. The method of completing squares is a bit difficult as compared to the direct formula for the root of quadratic equations. Students must remember both of these rules