Question

By quadratic equation solve
√3x^2 - 2x + 1/2 = 0

Answer 1

Answer:

yes, it is a quadratic equation.

Answer 2

Answer:

Required solution of the given quadratic equation is $x = \frac{ -2 \pm\sqrt{2( 2 - \sqrt{3} ) } }{2 \sqrt{3} }$

Step-by-step explanation:

Given,

$\sqrt{3} {x}^{2} - 2x + \frac{1}{2} = 0....(1)$

We are comparing equation (1) with

$a {x}^{2} + bx + c = 0$

and get

$a = \sqrt{3} \\ b = - 2 \\ c = \frac{1}{2}$

We know by Sridhar Acharya theorem,

Solution of

$a {x}^{2} + bx + c = 0$ is $x = \frac{ - b \pm\sqrt{ {b}^{2} - 4ac } }{2a}$

So, solution of equation (1) is

$x = \frac{ - ( - 2) \pm\sqrt{ {( - 2)}^{2} - 4 \sqrt{3} \times \frac{1}{2} } }{2 \sqrt{3} } \\ x = \frac{ -2 \pm\sqrt{ 4 - 2 \sqrt{3} } }{2 \sqrt{3} } \\ x = \frac{ -2 \pm\sqrt{ 2(2 - \sqrt{3}) } }{2 \sqrt{3} }$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549