Question

Find the zeroes of 4√3x^2 + 4√3x - 3√3, and also verify it by sum of zeroes and product of zeroes.​

Answer 1

Answer:

Given :-

• 4√3x² + 4√3x - 3√3

To Find :-

• What are the zeros of quadratic polynomial and verify it by sum of zeroes and product of zeroes.

Solution :-

Given Equation :

$\bigstar\: \: \bf{4\sqrt{3}x^2 + 4\sqrt{3}x - 3\sqrt{3}}$

Let,

$\implies \sf p(x) =\: 4\sqrt{3}x^2 + 4\sqrt{3}x - 3\sqrt{3}$

Zero of the polynomial is the value of x where p(x) = 0

By putting p(x) = 0 we get,

$\implies \sf 4\sqrt{3}x^2 + 4\sqrt{3}x - 3\sqrt{3} =\: 0$

$\implies \sf 4\sqrt{3}x^2 + (6\sqrt{3} - 2\sqrt{3})x - 3\sqrt{3} =\: 0$

$\implies \sf 4\sqrt{3}x^2 + 6\sqrt{3}x - 2\sqrt{3}x - 3\sqrt{3}\: \: \bigg\lgroup \sf\bold{\pink{By\: splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf 2\sqrt{3}x(2x + 3) - \sqrt{3}(2x + 3) =\: 0$

$\implies \sf (2\sqrt{3}x - \sqrt{3})(2x + 3) =\: 0$

$\longrightarrow \bf 2\sqrt{3}x - \sqrt{3} =\: 0$

$\longrightarrow \sf 2\sqrt{3}x =\: \sqrt{3}$

$\longrightarrow \sf x =\: \dfrac{\cancel{\sqrt{3}}}{2\cancel{\sqrt{3}}}$

$\longrightarrow \sf\bold{\red{x =\: \dfrac{1}{2}}}$

$\longrightarrow \bf 2x + 3 =\: 0$

$\longrightarrow \sf 2x =\: - 3$

$\longrightarrow \sf\bold{\red{x =\: \dfrac{- 3}{2}}}$

${\small{\bold{\underline{\therefore\: The\: zeroes\: of\: quadratic\: polynomial\: is\: \dfrac{1}{2}\: and\: \dfrac{- 3}{2}\: .}}}}$

Hence, we get :

• α = 1/2
• β = - 3/2

Now, we have to verify it by sum of zeroes and product of zeroes :

Given Equation :

$\bigstar\: \: \bf{4\sqrt{3}x^2 + 4\sqrt{3}x - 3\sqrt{3}}$

By comparing with ax² + bx + c we get,

• a = 4√3
• b = 4√3
• c = - 3√3

❒ Sum Of Zeroes :

As we know that :

$\clubsuit$ Sum Of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Sum\: Of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

We have :

• a = 4√3
• b = 4√3
• α = 1/2
• β = - 3/2

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{1}{2} + \dfrac{- 3}{2} =\: \dfrac{- 4\sqrt{3}}{4\sqrt{3}}$

$\longrightarrow \sf \dfrac{1 + (- 3)}{2} =\: \dfrac{- 4\sqrt{3}}{4\sqrt{3}}$

$\longrightarrow \sf \dfrac{1 - 3}{2} =\: \dfrac{- 4\sqrt{3}}{4\sqrt{3}}$

$\longrightarrow \sf \dfrac{- \cancel{2}}{\cancel{2}} =\: \dfrac{- \cancel{4\sqrt{3}}}{\cancel{4\sqrt{3}}}$

$\longrightarrow \sf\bold{\purple{- 1 =\: - 1}}$

Hence, Verified.

❒ Product Of Zeroes :

As we know that :

$\clubsuit$ Product Of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Product\: Of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

We have :

• a = 4√3
• c = - 3√3
• α = 1/2
• β = - 3/2

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{1}{2} \times \dfrac{- 3}{2} =\: \dfrac{- 3\cancel{\sqrt{3}}}{4\cancel{\sqrt{3}}}$

$\longrightarrow \sf \bold{\purple{\dfrac{- 3}{4} =\: \dfrac{- 3}{4}}}$

Hence, Verified.