Question

Find $k$ for which the equation $9x^2-kx+25=0$ has real and equal roots.

Answer 1

Step 1: Determining the discriminant

For the given quadratic equation $9x^2-kx+25=0$, the values of $a=9,b=-k$ and $c=25$.

Substituting the values in the discriminant formula $\left(D\right)=b^2-4ac$, we get

$$\begin{gathered} \Rightarrow D={(-k)}^2-4\left(9\right)\left(25\right) \\ \Rightarrow D=k^2-900 \end{gathered}$$

Step 2: Determining the values of $k$

Using the concept of nature of roots which says that when the roots of the quadratic equation are real and equal, $D=0$.
$$\begin{gathered} k^2-900=0 \\ \Rightarrow k^2-{\left(30\right)}^2=0 \\ \Rightarrow (k-30)(k+30)=0 \\ \therefore k=30,-30 \end{gathered}$$

Hence, the values of $k$ are $-30$ and $30$.