For a quadratic equation ax² + bx + c =0,
the term b² - 4ac is called discriminant (D) of
the quadratic equation
because it determines whether the quadratic equation has real roots or not ( nature of roots).
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
i)
x= (6-√12) / 2×2= 6-√4×3 /4 = 6 - 2√3 /4 = 2( 3 - √3) 4 = 3 - √3 /2
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
