Question

Write the discriminant of ${(a+b)}^2{x}^2+8(a^2-b^2)x+16{(a-b)}^2=0$ and determine the nature of its root.

Answer 1

Step 1: Comparison with the standard form of quadratic equation $A{x}^2+Bx+C=0$

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

Here, $A={(a+b)}^2,B=8(a^2-b^2)$ and $C=16{(a-b)}^2$.

Step 2: Finding the discriminant

Using the formula for discriminant $\left(D\right)=B^2-4AC$ and substituting the values of $A,B$ and $C$ we get,

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & D={\left[8\right(a^2-b^2\left)\right]}^2-4\left(16\right){(a+b)}^2{(a-b)}^2\\ & & ={\left[8\right(a^2-b^2\left)\right]}^2-64{(a+b)}^2{(a-b)}^2 \\ ={\left[8\right(a^2-b^2\left)\right]}^2-\left(8^2\right){(a+b)}^2{(a-b)}^2 \\ ={\left[8\right(a^2-b^2\left)\right]}^2-{\left[8(a+b)(a-b)\right]}^2 \\ ={\left[8\right(a^2-b^2\left)\right]}^2-{\left[8(a^2-b^2)\right]}^2 \\ =0\end{array} \end{gathered}$$

Step 3: Determining the nature of the roots of the equation

We know that the value of the discriminant decides the nature of the roots of the quadratic equation.

If $D>0$, then the quadratic equation has real and unequal roots.

If $D=0$, then the quadratic equation has real and equal roots.

If $D<0$, then the quadratic equation has no real roots.

For the given quadratic equation, $D=0$. So, the equation has real and equal roots.

Therefore, the quadratic equation ${(a+b)}^2{x}^2+8(a^2-b^2)x+16{(a-b)}^2=0$ has real and equal roots.