Question

whether (-√3) is a real root?​

Answer 1

i) x² – 3x + 5 = 0

On Comparing it with ax² + bx + c = 0, we get

Here, a = 2, b = -3 and c = 5

Discriminant (D) = b² – 4ac

D = (- 3)² - 4 (2) (5)  

D = 9 – 40

D= – 31<0

As b² – 4ac < 0,

Hence, no real root is possible .

(ii) 3x² – 4√3x + 4 = 0

On Comparing it with ax² + bx + c = 0, we get

Here,a = 3, b = -4√3 and c = 4

Discriminant(D) = b² – 4ac

D= (-4√3)2 – 4(3)(4)

D= 48 – 48 = 0

As b² – 4ac = 0,

Hence,  real roots exist & they are equal to each other.

roots will be –b/2a and –b/2a.

-b/2a = -(-4√3)/2×3  

= 4√3/6  

= 2√3/3  

multiplying the numerator & denominator by √3

= (2√3) (√3) / (3)(√3)  

= 2 ×3 / 3 ×√3  

= 2/√3

Hence , the equal roots are 2/√3 and 2/√3.

(iii) 2x² – 6x + 3 = 0

On Comparing this equation with ax² + bx + c = 0, we get

Here,a = 2, b = -6, c = 3

Discriminant (D)= b² – 4ac

D= (-6)2 – 4 (2) (3)

D= 36 – 24 = 12

As b2 – 4ac > 0,

Hence, two distinct real roots exist for this equation

x= -b/2a + √D/2a  &     x= -b/2a - √D/2a

x= (6+√12) / 2×2

x= 6+√4×3 /4  

x= 6 + 2√3 /4  

x= 2( 3 + √3) 4  

x = 3 + √3 /2

x= (6-√12) / 2×2

x= 6-√4×3 /4  

x= 6 - 2√3 /4  

x= 2(3 - √3)/ 4  

x= 3 - √3 /2

Hence the real roots are 3 + √3 /2   & 3 -√3 /2