Question

define Find the roots of the quadratic equation √3x² - 2x -√3-0​

Answer 1

Answer:

x = √3 or -1/√3

Step-by-step explanation:

√3 x² - 2x - √3 = 0

√3 x² - 3x + 1x - √3 = 0

√3 x (x - √3) + 1 (x - √3)

(√3 x + 1)(x - √3)

x = √3 or -1/√3

Answer 2

Step-by-step explanation:

Given :-

√3x² - 2x -√3 = 0

To find:-

Find the roots ?

Solution:-

Factorization method :-

Given equation is √3x² - 2x -√3= 0

=> √3 x² -3x+x -√3 = 0

=> √3x(x-√3) +1(x-√3) = 0

=> (x-√3)(√3x+1) = 0

=> x-√3 = 0 or √3x+1 = 0

=> x = √3 or √3x = -1

=> x = √3 or x = -1/√3

Roots are √3 and -1/√3

Completing the square method :-

Given equation is √3x² - 2x -√3= 0

On dividing by √3 both sides then

=> x²-(2/√3)x-1 = 0

=> x²-2(x)(1/√3)-1 = 0

=> x²-2(x)(1/√3) = 1

On adding (1/√3)² both sides then

=> x²-2(x)(1/√3)+(1/√3)² = 1+(1/√3)²

=> [x-(1/√3)]² = 1+(1/3)

=> [x-(1/√3)]² =(3+1)/3

=> [x-(1/√3)]² = 4/3

=> x-(1/√3) =±√(4/3)

=> x-(1/√3) =±2/√3

=> x = (±2/√3)+(1/√3)

=> x = (1/√3)+(2/√3) or (1/√3)-(2/√3)

=>x = (1+2)/√3 or (1-2)/√3

=> x = 3/√3 or -1/√3

=> x = (√3×√3)/√3 or -1/√3

=> x = √3 or -1/√3

Roots are √3 and -1/√3

Formula method:-

Given equation is √3x² - 2x -√3= 0

On Comparing this with the standard quadratic equation ax²+bx+c = 0

a =√3

b = -2

c = -√3

Quadratic formula x = [-b±√(b²-4ac)]/2a

=> x = [-(-2)±√{(-2)²-4(√3)(-√3)]/2(√3)

=> x = [2±√(4+12)]/2√3

=> x = (2±√16)/2√3

=> x = (2±4)/2√3

=> x = 2(1±2)/2√3

=> x = (1±2)/√3

=> x = (1+2)/√3 or (1-2)/√3

=> x = 3/√3 or -1/√3

=> x = (√3×√3)/√3 or -1/√3

=> x = √3 or -1/√3

Roots are √3 and -1/√3

Answer:-

The roots of the given equation are √3 and -1/√3

Used formulae:-

• (a-b)² = a²-2ab+b²
• x = [-b±√(b²-4ac)]/2a
• The standard quadratic equation ax²+bx+c = 0

Used Methods:-

• Factorization method (Splitting the middle term)
• Completing the square method
• Formula method (Sridharacharya formula)