Question

18. x2 - (√3 + 1)x + √3 = 0​

Answer 1

Answer:

1

Step-by-step explanation:

x² - (√3 + 1)x + √3 = 0

x² - 2(√3 + 1)(x)/2 + √3 = 0

x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0

[x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0

[x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4

[x - (√3 + 1)/2]² = (√3 - 1)²/2²

[x - (√3 + 1)/2] = (√3 - 1)/2

taking (+ve)

x = (√3 - 1)/2 + (√3 + 1)/2

x = (√3 - 1 + √3 + 1)/2

x = 2√3/2 = √3

taking (-ve)

x = -(√3 - 1)/2 + (√3 + 1)/2

x = (-√3 + 1 + √3 + 1)/2

x = 2/2 = 1

PLEASE MARK AS BRAINLIEST

Answer 2

We know that,

$p(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ where \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: p(x)$

So here,

$\alpha + \beta = \sqrt{3} + 1$

$\alpha \beta = \sqrt{3}$