# Find the zeros of 6x^2-3?

Feb 24, 2018

Zeros are $\frac{1}{\sqrt{2}}$ and $- \frac{1}{\sqrt{2}}$

#### Explanation:

Here in $6 {x}^{2} - 3$, there is no middle term and polynomial appears as difference of remaining two terms. In such cases, we can write it as ${a}^{2} - {b}^{2}$ and factors are $\left(a + b\right) \left(a - b\right)$, while zeros can be obtained by putting $a + b = 0$ and $a - b = 0$.

Here, however, one desires to use the method of splitting the middle term and for this one can add and subtract $a b$. Here as ${a}^{2} = 6 {x}^{2}$, our $a = x \sqrt{6}$ and as $3 = {\left(\sqrt{3}\right)}^{2}$, we add and subtract $x \sqrt{6} \times \sqrt{3} = \sqrt{18} x$ and then

$6 {x}^{2} - 3$

= $6 {x}^{2} + \sqrt{18} x - \sqrt{18} x - 3$

= $\sqrt{6} x \left(\sqrt{6} x + \sqrt{3}\right) - \sqrt{3} \left(\sqrt{6} x + \sqrt{3}\right)$

= $\left(\sqrt{6} x - \sqrt{3}\right) \left(\sqrt{6} x + \sqrt{3}\right)$

Here as middle term is $0$, we have split it into $+ \sqrt{18} x - \sqrt{18} x$

and zeros are given by $\sqrt{6} x - \sqrt{3} = 0$ i.e. $x = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$

and $\sqrt{6} x + \sqrt{3} = 0$ i.e. $x = - \frac{\sqrt{3}}{\sqrt{6}} = - \frac{1}{\sqrt{2}}$