(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution
(i) 2x2 – 3x + 5 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
We know, Discriminant = b2 – 4ac
= ( – 3)2 – 4 (2) (5) = 9 – 40
= – 31
As you can see, b2 – 4ac < 0
Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
We know, Discriminant = b2 – 4ac
= (-4√3)2 – 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac = 0,
Real roots exist for the given equation, and they are equal to each other.
Hence, the roots will be –b/2a and –b/2a.
–b/2a = -(-4√3)/2 × 3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.
(iii) 2x2 – 6x + 3 = 0
Comparing the equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
As we know, Discriminant = b2 – 4ac
= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac > 0,
Therefore, there are distinct real roots that exist for this equation, 2x2 – 6x + 3 = 0
x = [-b ± √(b2 – 4ac)]/2a
= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)
= (6 ± 2√3 )/4
= (3 ± √3)/2
Therefore, the roots for the given equation are (3 + √3)/2 and (3 – √3)/2