D= b² - 4ac

So a quadratic equation ax² + bx + c =0, has

i) Two distinct real roots, if b² - 4ac >0 , then

**x= -b/2a + √D/2a**

**&**

**x= -b/2a - √D/2a**

ii) Two equal real roots, if b² - 4ac = 0 , then **x= -b/2a or -b/2a**

iii) No real roots, if b² - 4ac <0

Solution:i)

x² – 3x + 5 = 0

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

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

Discriminant (D) = b² – 4ac

⇒ ( – 3)2 – 4 (2) (5) = 9 – 40

⇒ – 31<0

As b2 – 4ac < 0,

Hence, no real root is possible .

(ii) 3x² – 4√3x + 4 = 0

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

a = 3, b = -4√3 and c = 4

Discriminant(D) = b² – 4ac

⇒ (-4√3)2 – 4(3)(4)

⇒ 48 – 48 = 0

As b² – 4ac = 0,

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

the 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

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

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

Discriminant (D)= b² – 4ac

= (-6)2 – 4 (2) (3)

= 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= 6+√4×3 /4 = 6 + 2√3 /4 = 2( 3 + √3) 4 = 3 + √3 /2

**x = 3 + √3 /2**

x=

**(6-**√12) / 2×2= 6-√4×3 /4 = 6 - 2√3 /4 = 2( 3 - √3) 4 = 3 - √3 /2

**x= 3 - √3 /2**

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

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