Question

please solve this question

Answer 1

We need to find the zeroes of 6(x^2)-3...
.°. 6(x^2)-3 = 0
3{2(x^2)-1} = 0
2(x^2)-1 = 0
2(x^2) = 1
(x^2) = 1/2
.°. x = 1/2 or -1/2
.°. 1/2 and -1/2 are the zeroes of the polynomial!

Answer 2

Question:

Find the zeroes of the quadratic polynomial f(x) = $6{x}^{2} - 3$, and verify the relationship between the zeroes and its coefficients.

Solution:

To find the zeroes

f(x) = 0

$f(x) = 6{x}^{2} - 3 = 0 \\ \implies {x}^{2} = \dfrac{3}{6} \\ \implies {x}^{2} = \dfrac{1}{2} \\ \implies x = \pm \sqrt{\dfrac{1}{2}} \\ \implies x = \pm \dfrac{1}{\sqrt{2}}$

So, the zeroes are $-\dfrac{1}{\sqrt{2}}$ and $\dfrac{1}{\sqrt{2}}$.

To verify the relationship

Sum of zeroes = $-\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}$

= 0

Product of zeroes = $-\dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}}$

$= \dfrac{-1}{2}$

Sum of zeroes = $\dfrac{-Coefficient\: of \: x}{Coefficient \: of \: {x}^{2}}$

$= \dfrac{-(0)}{6} \\ = 0$

Product of zeroes = $\dfrac{Constant \: term}{Coefficient\: of \: {x}^{2}}$

$= \dfrac{-3}{6} \\ = \dfrac{-1}{2}$

Hence, verified.