Question

solve for x if x^2-(√3+1)x+√3=0, find the roots through quadratic formula or sridhara's formula.

Answer 1

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Here is your answer...✌

Given quadratic equation :

x² - (√3 + 1)x + √3 = 0

⇒ a = 1 , b = -(√3 + 1) , c = √3

Discriminant , D = b² - 4ac

D = (√3 + 1)² - 4×1×√3

D = (√3)² + 1² + 2√3 - 4√3

D = (√3)² + 1² - 2√3

D = (√3 - 1)²

Now using quadratic formula,
we get,

x = ( -b ± √D ) / 2a

x = [ √3 + 1 ± ( √3 - 1 ) ] / 2×1

x = (√3+1+√3-1)/2 or x = (√3+1-√3+1)/2

⇒ x = 2√3 / 2 or x = 2/2

⇒ x = √3 or x = 1

Answer 2

Answer:

Step-by-step explanation:

x² - (√3 + 1)x + √3 = 0

⇒ a = 1 , b = -(√3 + 1) , c = √3

Discriminant , D = b² - 4ac

D = (√3 + 1)² - 4×1×√3

D = (√3)² + 1² + 2√3 - 4√3

D = (√3)² + 1² - 2√3

D = (√3 - 1)²

Now using quadratic formula,

we get,

x = ( -b ± √D ) / 2a

x = [ √3 + 1 ± ( √3 - 1 ) ] / 2×1

x = (√3+1+√3-1)/2 or x = (√3+1-√3+1)/2

⇒ x = 2√3 / 2 or x = 2/2

⇒ x = √3 or x = 1