solve the equation x^2-( √3+1)x+ √3x=0 using quadratic formula
Answers
Answered by
1
Comparing the equation with the general formula, we get : a = 1, b = -√3-1, c = √3
∆ = b²-4ac
= (-√3-1)²- ( 4 × 1 × √3 )
= [ 3 - ( 2 × (-√3) × ( -1 ) ) + 1 ] - 4√3
= [ 3 - 2√3 + 1 ] - 4√3
= 4 - 2√3 - 4√3
= 4 - 6√3
We know,
x = (-b±√b²-4ac)/2a
= (√3+1±√4-6√3)/2
= (√3+1+√4-6√3)/2 or (√3+1-√4-6√3)/2
.....✌
∆ = b²-4ac
= (-√3-1)²- ( 4 × 1 × √3 )
= [ 3 - ( 2 × (-√3) × ( -1 ) ) + 1 ] - 4√3
= [ 3 - 2√3 + 1 ] - 4√3
= 4 - 2√3 - 4√3
= 4 - 6√3
We know,
x = (-b±√b²-4ac)/2a
= (√3+1±√4-6√3)/2
= (√3+1+√4-6√3)/2 or (√3+1-√4-6√3)/2
.....✌
Similar questions