Question

Find the value of 'k' for which the following set of quadratic equation has exactly one common root
$x^2-kx+10=0$ and $x^2+kx-18=0.$

• A. $\pm 3$
• B. $\pm 7$
• C. $\pm 9$
• D. $\pm 11$

Answer 1

Answer: (B)

$\pm 7$

$x^2-kx+10=0$

$x^2+kx-18=0$

Let the common root be a

$a^2-ka+10=0$

$a^2+ka-18=0$

_________________

$2a^2-8=0$

$2a^2=8$

$a^2=4$

$a=\pm 2$

$\bullet a=+2$

$4-2k+10=0$

$2k=14$

$k=7$

$k=\pm 7$

$\bullet a=-2$

$4+2k+10=0$

$2k=-14$

$k=-7$.