Question

If 1 is a root of the quadratic equation $3x^2+ax-2=0$ and the quadratic equation $a(x^2+6x)-b=0$ has equal roots, find the value of b.

Answer 1

The given quadratic equation is $3x^2+ax-2=0$, and one root is 1.

Then, it satisfies the given equation.

$$\begin{gathered} 3{\left(1\right)}^2+a\left(1\right)-2=0 \\ \Rightarrow 3+a-2=0 \\ \Rightarrow 1+a=0 \\ \Rightarrow a=-1 \end{gathered}$$

The quadratic equation $a(x^2+6x)-b=0$, has equal roots.

Putting the value of a, we get

$$\begin{gathered} -1\left(x^2+6x\right)-b=0 \\ \Rightarrow x^2+6x+b=0 \end{gathered}$$

Here, $A=1,B=6\mathrm{and}C=b$.

As we know that $D=B^2-4AC$

Putting the values of $A=1,B=6\mathrm{and}C=b$.

$$\begin{gathered} D={\left(6\right)}^2-4\left(1\right)\left(b\right) \\ =36-4b \end{gathered}$$

The given equation will have real and equal roots, if D = 0

Thus, $36-4b=0$
$$\begin{gathered} \Rightarrow 4b=36 \\ \Rightarrow b=9 \end{gathered}$$

Therefore, the value of b is $9$.