Question:
The zeros of the polynomial $4 x^{2}+5 \sqrt{2} x-3$ are
(a) $-3 \sqrt{2}, \sqrt{2}$
(b) $-3 \sqrt{2}, \frac{\sqrt{2}}{2}$
(c) $\frac{-3 \sqrt{2}}{2}, \frac{\sqrt{2}}{4}$
(d) none of these
Solution:
(c) $-\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$
Let $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$
$=>4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$
$=>2 \sqrt{2} x(\sqrt{2} x+3)-1(\sqrt{2} x+3)=0$
$=>(\sqrt{2} x+3)(2 \sqrt{2} x-1)=0$
$=>x=-\frac{3}{\sqrt{2}}$ or $x=\frac{1}{2 \sqrt{2}}$
$=>x=-\frac{3}{\sqrt{2}}$ or $x=\frac{1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$