Question

find the zeros of the quadratic equation f(x)= 6x²-3, and verify the relation between its zeros and coefficients.
using the formula
alpha+beta= -b/a
alpha × beta = c/a​

Answer 1

Answer:

$f(x) = 6 {x}^{2} - 3 \\ = > 3(2 {x}^{2} - 1) = 0 \\ = > 2 {x}^{2} - 1 = 0 \\ = > 2 {x}^{2} = 1 \\ = > {x}^{2} = \frac{1}{2} \\ = > x = ±\frac{1}{ \sqrt{2} } \\ \alpha = \frac{1}{ \sqrt{2} } and \: \beta = - \frac{1}{ \sqrt{2} } \\ now, \\ = > \alpha + \beta \\ = \frac{1}{ \sqrt{2} } + \frac{ - 1}{ \sqrt{2} } \\ = \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \\ = 0 \\ and \: \frac{ - b}{a} = \frac{0}{6} \\ = 0 \\ so, \: \alpha + \beta = \frac{ - b}{a} \\= > \alpha \beta = \frac{1}{ \sqrt{2} } \times \frac{ - 1}{ \sqrt{2} } \\ = - \frac{1}{2} \\ and \: \frac{c}{a} = \frac{ - 3}{6} \\ = \frac{ - 1}{2} \\ so, \: \alpha \beta = \frac{c}{a}$

Answer 2

Answer:

Step-by-step explanation:

We are already given the following formulas :-

$\alpha + \beta = \dfrac{-b}{a}$

$\alpha \times \beta = \dfrac{c}{a}$

Given that the polynomial is :-

$\sf{6x^2 + 0x - 3}$

Now, we will use factorisation as our first weapon :-

6x² -3 = 0 [As we have to find the zeroes of the equation]

6x² = 3

x² = $\dfrac{3}{6}$

x² = $\dfrac{1}{2}$

x = $\dfrac{1}{\sqrt{2}}$

Now, we can take one zero as 1/Root 2 and other -1/root 2

$\dfrac{1}{\sqrt{2}} + \dfrac{-1}{\sqrt{2}} | \dfrac{0}{6} = 0$

$0 |0$ Hence verified.

$\dfrac{1}{\sqrt{2}} \times \dfrac{-1}{\sqrt{2}} | \dfrac{-3}{6} = 0$

$\dfrac{1}{2} | \dfrac{1}{2}$

Hence, proved.