Question

FIND ZERO'S OF POLYNOMIAL AND VERIFY THE EQUATION
$2 \sqrt{3}x {}^{2} - 5x + 3$
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Answer 1

Correct Question:

Find zeroes of the polynomial and verify the equation, P(x) = 2√3x² - 5x + √3.

Answer:

• Zeroes of the polynomial are 1/√3 and √3/2.

Step-by-step explanation:

Given polynomial:

• 2√3x² - 5x + √3

To Find:

• Zeroes of the polynomial and verify the equation.

Finding the zeroes of the polynomial:

⟿ 2√3x² - 5x + √3 = 0

Splitting 5x.

⟿ 2√3x² - 3x - 2x + √3 = 0

Taking common.

⟿ √3x(2x - √3) - 1(2x - √3) = 0

⟿ (√3x - 1) (2x - √3) = 0

⟿ x = 1/√3 or x = √3/2

∴ Zeroes of the polynomial are 1/√3 and √3/2.

Verifying the equation:

P(x) = 0

P(1/√3) = 2√3x² - 5x + √3

P(1/√3) = 2√3•(1/√3)² - 5•1/√3 + √3

P(1/√3) = 2√3/3 - 5/√3 + √3

P(1/√3) = 2/√3 - 5/√3 + 3/√3

P(1/√3) = 0

Here, verified.

P(x) = 0

P(√3/2) = 2√3x² - 5x + √3

P(√3/2) = 2√3•(√3/2)² - 5•√3/2 + √3

P(√3/2) = 2√3•3/4 - 5√3/2 + √3

P(√3/2) = 6√3/4 - 10√3/4 + 4√3/4

P(√3/2) = 0

Here also, verified.

Answer 2

Given :-

$\sf 2\sqrt{3}x^{2} -5x+\sqrt3$

To Find :-

Zeroes of polynomial

Solution :-

$\sf 2\sqrt{3}x^{2} -5x+\sqrt3$

$\sf 2\sqrt{3}x^{2} - (3x + 2x) + \sqrt3$

$\sf 2\sqrt{3}x^{2} - 3x-2x+\sqrt3$

$\sf \sqrt{3}x(2x - \sqrt{3}) - 1(2x - \sqrt{3}) = 0$

$\sf Either$

x = 1/√3

or

x = √3/2

Verification

$\sf 2 \sqrt{3} \times \bigg(\dfrac{1}{\sqrt{3}}\bigg)^2- 5 \times \dfrac{1}{\sqrt{3}} + \sqrt{3}$

$\sf 2 \sqrt{3} \times \bigg(\dfrac{1}{3}\bigg)- 5 \times \dfrac{1}{\sqrt{3}} + \sqrt{3}$

$\sf 2 \sqrt{3} \times \bigg(\dfrac{1}{3}\bigg)- \dfrac{5}{\sqrt{3}} + \sqrt{3}$

$\sf \dfrac{2\sqrt{3}}{3} - \dfrac{5}{\sqrt{3}}+\sqrt 3$

$\sf = \dfrac{2}{\sqrt{3}} - 5$

0

Again

$\sf 2\sqrt{3} \times \bigg(\dfrac{\sqrt{3}}{2}\bigg)^2 - 5 \times \dfrac{\sqrt{3}}{2} + \sqrt3$

$\sf 2\sqrt{3} \times \bigg(\dfrac{3}{4}\bigg) - 5 \times \dfrac{\sqrt{3}}{2} + \sqrt3$

$\sf = 0$