Question

Find the positive value(s) of k for which quadratic equations x² + kx + 64 = 0 and x² – 8x + k = 0 both will have real roots ?

Answer 1

The given quadratic equations are \[x^2 + kx + 64 = 0\] and \[x^2 - 8x + k = 0\]

\[x^2 + kx + 64 = 0\] has real roots.

\[\therefore D = k^2 - 4 \times 1 \times 64 \geq 0\]
\[ \Rightarrow k^2 - {16}^2 \geq 0\]
\[ \Rightarrow \left( k + 16 \right)\left( k - 16 \right) \geq 0\]

Case I

\[k + 16 \geq 0\ \text{and} \ k - 16 \geq 0\]
\[ \Rightarrow k \geq - 16 \ \text{and} \ k \geq 16\]
\[ \Rightarrow k \geq 16 . . . . . \left( 1 \right)\]

Case 2

\[k + 16 \leq 0\ \text{and}\ k - 16 \leq 0\]
\[ \Rightarrow k \leq - 16\ \text{and}\ k \leq 16\]
\[ \Rightarrow k \leq - 16 . . . . . \left( 2 \right)\]

From (1) and (2), we get

\[k \in ( - \infty , - 16] \cup [16, \infty )\]

Now,

\[x^2 - 8x + k = 0\] has real roots.

\[\therefore D = \left( - 8 \right)^2 - 4 \times 1 \times k \geq 0\]
\[ \Rightarrow 64 - 4k \geq 0\]
\[ \Rightarrow - 4k \geq - 64\]
\[ \Rightarrow k \leq \frac{- 64}{- 4}\]
\[ \Rightarrow k \leq 16 \]
\[ \Rightarrow k \in ( - \infty , 16] . . . . . \left( 4 \right)\]

From (3) and (4), we get

\[k \in ( - \infty , - 16] \cup \left\{ 16 \right\}\]

Since k is positive, it is 16.
Thus, the positive value of k for which both the equations will have real roots is 16.