Question

Solve for the value of ‘x’: $4{{x}^{2}}-4ax+\left( {{a}^{2}}-{{b}^{2}} \right)=0$.

Answer 1

Hint: Try to observe the given equation and factorize the given equation in the form of $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$, and hence, find the values of ‘x’ with the help of that equation.

Here, we have quadratic equation in ’x’ as
$4{{x}^{2}}-4ax+\left( {{a}^{2}}-{{b}^{2}} \right)=0$ ……………. (i)
We can rewrite the above equation as
$\left( 4{{x}^{2}}-4ax+{{a}^{2}} \right)-{{b}^{2}}=0$
Now, observe the terms of the bracket, and factorize it; we can rewrite the terms of bracket as
$\left( {{\left( 2x \right)}^{2}}-2\times a\times 2x+{{\left( a \right)}^{2}} \right)-{{b}^{2}}=0$ ………….. (ii)
We know the algebraic identity
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & \Rightarrow {{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}...........(iii) \\
\end{align}$
Now, compare the terms of the bracket with the equation (iii), so, we can simplify the equation (ii) as
${{\left( 2x-a \right)}^{2}}-{{b}^{2}}=0$……………. (iv)
Now, we can use the algebraic identity given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$……………….. (v)
So, applying the above identity with the equation (iv), hence, we get
$\left( 2x-a-b \right)\left( 2x-a+b \right)=0$……………. (vi)
Now, we can observe that expression xy = 0 if, x = 0 or y = 0. So, we can get values of ‘x’ from equation (vi) by putting $\left( 2x-a-b \right)$ and $\left( 2x-a+b \right)$ to 0,
Hence, we get

$\begin{align}
  & 2xab=0 \\
 & \Rightarrow 2x\text{ }=\text{ }a\text{ }+\text{ }b \\
 & \Rightarrow x=\dfrac{a+b}{2} \\
\end{align}$
And we can put 2x –a + b to 0 as well. Hence, we get
$\begin{align}
  & \begin{array}{*{35}{l}}
   2xa+b=0 \\
   \Rightarrow 2x=ab \\
\end{array} \\
 & \Rightarrow x=\dfrac{a-b}{2} \\
\end{align}$
Hence, values of ‘x’ from the quadratic given in the problem are $\dfrac{a+b}{2}$and $\dfrac{a-b}{2}$
Hence, answer is $x=\dfrac{a-b}{2}$, $\dfrac{a+b}{2}$

Note: Another approach for the given question would be that we can solve the quadratic by using the quadratic formula which is given as:
Roots of quadratic $A{{x}^{2}}+Bx+C=0$ are
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
Hence, values of ’x’ from the given quadratic are given as
$x=\dfrac{4a\pm \sqrt{{{\left( 4a \right)}^{2}}-4\times 4\left( {{a}^{2}}-{{b}^{2}} \right)}}{8}$
Observing $4{{x}^{2}}-4ax+{{a}^{2}}$ to ${{\left( x-2a \right)}^{2}}$ is the key point of the question with respect to the solution.