Question

If −5 is a root of the quadratic equation $2x^2+px-15=0$ and the quadratic equation $p(x^2+x)+k=0$ has equal roots, find the value of k.

Answer 1

The given quadratic equation is $2x^2+px-15=0$, and one root is −5.

Then, it satisfies the given equation.

$$\begin{gathered} 2{\left(-5\right)}^2+p\left(-5\right)-15=0 \\ \Rightarrow 50-5p-15=0 \\ \Rightarrow -5p=-35 \\ \Rightarrow p=7 \end{gathered}$$

The quadratic equation $p(x^2+x)+k=0$, has equal roots.

Putting the value of p, we get

$$\begin{gathered} 7\left(x^2+x\right)+k=0 \\ \Rightarrow 7x^2+7x+k=0 \end{gathered}$$

Here, $a=7,b=7\mathrm{and}c=k$.

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

Putting the values of $a=7,b=7\mathrm{and}c=k$.

$$\begin{gathered} D={\left(7\right)}^2-4\left(7\right)\left(k\right) \\ =49-28k \end{gathered}$$

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

Thus, $49-28k=0$
$$\begin{gathered} \Rightarrow 28k=49 \\ \Rightarrow k=\frac{49}{28} \\ \Rightarrow k=\frac{7}{4} \end{gathered}$$

Therefore, the value of k is $\frac{7}{4}$.