Question

Using the quadratic formula solve ${(a+b)}^2{x}^2+8(a^2-b^2)x+16{(a-b)}^2=0$ for $x$.

Answer 1

Step 1: Determining the discriminant

Given equation is ${(a+b)}^2{x}^2+8(a^2-b^2)x+16{(a-b)}^2=0$.

The discriminant formula is $D=B^2-4AC$ (Considering $A,B,C$ for ease of calculation because the question has variables $a,b,c$).

Where, for the given equation $A={(a+b)}^2,B=8(a^2-b^2)$ and $C=16{(a-b)}^2$.

Substituting the values in the discriminant formula

$$\begin{gathered} D=B^2-4AC \\ \Rightarrow D={\left[8\left(a^2-b^2\right)\right]}^2-4{\left(a+b\right)}^2\left[16{\left(a-b\right)}^2\right] \\ \Rightarrow D=64\left(a^4-2a^2{b}^2+b^4\right)-4\left(a^2+2ab+b^2\right)\left\{16\left(a^2-2ab+b^2\right)\right\} \\ \Rightarrow D=64a^4-128a^2{b}^2+64b^4-\left(4\right)\left(16\right)\left[a^2\left(a^2-2ab+b^2\right)+2ab\left(a^2-2ab+b^2\right)+b^2\left(a^2-2ab+b^2\right)\right] \\ \Rightarrow D=64a^4-128a^2{b}^2+64b^4-64\left[a^4-2a^{3}b+a^2{b}^2+2a^{2}b-4a^2{b}^2+2a{b}^3+a^2{b}^2-2a{b}^3+b^4\right] \\ \Rightarrow D=64a^4-128a^2{b}^2+64b^4-64\left[a^4-2a^2{b}^2+b^4\right] \\ \Rightarrow D=64a^4-128a^2{b}^2+64b^4-{64}^4+128a^2{b}^2+64b^4 \\ \therefore D=0 \end{gathered}$$

Step 2: Determining the value of $x$

The quadratic formula is $\begin{array}{rcl}x& =& \frac{-B\pm \sqrt{B^2-4AC}}{2A}\end{array}$.

Substituting the values in the quadratic formula

$$\begin{gathered} \Rightarrow x=\frac{-8(a^2-b^2)}{2{(a+b)}^2} \\ \Rightarrow x=\frac{-8\left(a+b\right)\left(a-b\right)}{2{(a+b)}^2} \\ \therefore x=\frac{-4(a-b)}{(a+b)} \end{gathered}$$

Hence, the value of $x$ is $\frac{-4(a-b)}{(a+b)}$.