Question

If 2 is a root of the quadratic equation $3x^2+px-8=0$ and the quadratic equation $4x^2-2px+k=0$ has equal roots, find the value of k.

Answer 1

The given quadratic equation is $3x^2+px-8=0$, and one root is 2.

Then, it satisfies the given equation.

$$\begin{gathered} 3{\left(2\right)}^2+p\left(2\right)-8=0 \\ \Rightarrow 12+2p-8=0 \\ \Rightarrow 2p=-4 \\ \Rightarrow p=-2 \end{gathered}$$

The quadratic equation $4x^2-2px+k=0$, has equal roots.

Putting the value of p, we get

$$\begin{gathered} 4x^2-2(-2)x+k=0 \\ \Rightarrow 4x^2+4x+k=0 \end{gathered}$$

Here, $a=4,b=4\mathrm{and}c=k$.

As we know that $D=b^2-4ac$

Putting the values of $a=4,b=4\mathrm{and}c=k$.

$$\begin{gathered} D={\left(4\right)}^2-4\left(4\right)\left(k\right) \\ =16-16k \end{gathered}$$

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

Thus, $16-16k=0$
$$\begin{gathered} \Rightarrow 16k=16 \\ \Rightarrow k=1 \end{gathered}$$

Therefore, the value of k is 1.