Browse Questions

# If the numbers $\;1,x^2,6-x^2\;$ are in GP , then value of $x$ is :

$(a)\;\pm\sqrt{3}\qquad(b)\;\pm3\qquad(c)\;\pm\sqrt{2}\qquad(d)\;\pm2$

Can you answer this question?

Answer : (c) $\;\pm\sqrt{2}$
Explanation : For GP
$(x^2)^2=(1)\;(6-x^2)$
$x^4+x^2-6=0$
$(x^2+3)(x^2-2)=0$
$x^2=2$
$x=\pm\sqrt{2}.$
answered Jan 21, 2014 by