Question

$8x {}^{2} - \: 4$
Obtain zeroes of the polynomial

Answer 1

heya answer is here......

$8x {}^{2} - 4 = 0 \\ 4(2x {}^{2} - 1) = 0 \\ ( \sqrt{2} x + 1) \times ( \sqrt{2}x - 1) = 0 \\ x = \frac{ - 1}{ \sqrt{2} \: } or \: x = \frac{1}{ \sqrt{2} }$
it may be helps u dear.....

Answer 2

♥️Hi there ♥️

Kindly see attachment for 2 method.

$f(x) = 8 {x}^{2} - 4 \\ f(x) = ( \sqrt{8} x )^{2} - ( \sqrt{4} {}^{2})$
$f(x) = ( \sqrt{8} x - \sqrt{4} )( \sqrt{6} x + \sqrt{4} )$

$x = \frac{ \sqrt{4} }{ \sqrt{8} }$

It will cancel out

√1 / √2

$x = \frac{ - \sqrt{4} }{ \sqrt{8} }$

-√1 / √2

Thanks !!