Question

The value of $k(k>0)$ for which the equation $x^2+kx+64=0$ and $x^2-8x+k=0$ both will have real roots, is

• A. $8$
• B. $-16$
• C. $-64$
• D. $16$

Answer 1

Answer: (D)

Since $x^2+Kx+64=0$,
$x^2-8x+K=0$ have real roots,
$\therefore K^2-256\ge 0$ and $64-4K\ge 0$
$\Rightarrow K^2\ge 256$ and $16\ge K$
$\therefore K=16[\because K>0]$