Question

find the zeroes of 6x2-3​

Answer 1

$\: \: \: \: \: \: \: \: 6 {x}^{2} - 3 = 0 \\ = > \: \: \: \: \: \: \: 6 {x}^{2} = 3 \\ = > \: \: \: \: \: \: \: \: \: {x}^{2} = \frac{3}{6} \\ = > \: \: \: \: \: \: \: \: \: {x}^{2} = \frac{1}{2} \\ = > \: \: \: \: \: \: \: \: \: \: \: x = \sqrt{ \frac{1}{2} } \\ = > \: \: \: \: \: \: \: \: \: \: \: x = \: \: \frac{1}{ \sqrt{2} } \: \: \: \: or \: \: \: \: - \frac{1}{ \sqrt{2} }$

$So, \: the \: two \: zeroes \: are \frac{1}{ \sqrt{2} } \: and \: - \frac{1}{ \sqrt{2} }$

Answer 2

f(x) = 6x²-3

f(x) = 0

6x²-3 = 0

6x² = 3

x² = $\frac{3}{6}$

x² = $\frac{1}{2}$

x = $\frac{1}{2}$ OR $\frac{-1}{√2}$

Verification:

(i) 6($\frac{1}{2}$)² - 3

= 6($\frac{1}{2}$) - 3

= 3-3

= 0

(ii) 6($\frac{-1}{√2}$)² - 3

= 6(1/2) - 3

= 3-3

= 0

∴ The zeros of the polynomial 6x²-3 are $\frac{1}{2}$ and $\frac{1}{√2}$