Question

Using the quadratic formula, solve $9x^2-3(a^2+b^2)x+a^2{b}^2=0$ for $x$.

Answer 1

Step 1: Determining the discriminant

Given equation is $9x^2-3(a^2+b^2)x+a^2{b}^2=0$.

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 $A=9,B=-3\left(a^2+b^2\right)$ and $C=a^2{b}^2$.

Substituting the values in the discriminant formula

$$\begin{gathered} D=B^2-4AC \\ \Rightarrow D=(-3{(a^2+b^2))}^2-4\times 9\times a^2{b}^2 \\ \Rightarrow D=9\left[a^4+b^4+2a^2{b}^2-4a^2{b}^2\right] \\ \therefore D=9{\left(a^2-b^2\right)}^2 \end{gathered}$$

Step 2: Determining the values of $x$

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

Substituting the values $a,b$ and $D$ in the quadratic formula,

$$\begin{gathered} x=\frac{-B\pm \sqrt{B^2-4AC}}{2A} \\ \Rightarrow x=\frac{-\left[-3\left(a^2+b^2\right)\right]\pm \sqrt{9{\left(a^2-b^2\right)}^2}}{2\left(9\right)} \\ \Rightarrow x=\frac{3\left(a^2+b^2\right)\pm 3\left(a^2-b^2\right)}{18} \\ \Rightarrow x=\frac{\left(3a^2+3b^2\right)\pm \left(3a^2-3b^2\right)}{18} \\ \Rightarrow x=\frac{3a^2+3b^2+3a^2-3b^2}{18},\frac{3a^2+3b^2-3a^2+3b^2}{18} \\ \Rightarrow x=\frac{6a^2}{18},\frac{6b^2}{18} \\ \therefore x=\frac{a^2}{3},\frac{b^2}{3} \end{gathered}$$

Hence, the values of $x$ are $\frac{a^2}{3}$ and $\frac{b^2}{3}$.