Question

Find the zeros of the quadratic equation f(x)= 6x²-3, and verify the relation between its zeros and coefficients.

Answer 1

Answer:

Step-by-step explanation:

We are already given the following formulas :-

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

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

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.

Answer 2

Answer:

f(x) = 6x²-3

f(x) = 0

6x²-3 = 0

6x² = 3

x² = 3/6

x² = 1/2

x = 1/√2 , -1/√2

Roots of f(x) = 1/√2 , -1/√2

Verification = (i) 6(1/√2)² - 3

= 6(1/2) - 3

= 3-3

= 0

(ii) 6(-1/√2)² - 3

= 6(1/2) - 3

= 3-3

= 0