Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 - 3x + 5 = 0
(ii) 3x2 - 4√3x + 4 = 0
(iii) 2x2 - 6x + 3 = 0
(i) Consider the equation
x2 - 3x + 5 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= ( - 3)2 - 4 (2) (5) = 9 - 40
= - 31
As b2 - 4ac < 0,
Therefore, no real root is possible for the given equation.
(ii) 3x2 - 4√3x + 4 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 - 4ac
= (-4√3)2 - 4(3)(4)
= 48 - 48 = 0
As b2 - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And 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 this equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)2 - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation:
