Question 1:
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√3 x + 4 = 0
(III) 2x2 − 6x + 3 = 0
Answer:
We know that for a quadratic equation ax2 + bx + c = 0, discriminant is b2 − 4ac.
(A) If b2 − 4ac > 0 → two distinct real roots
(B) If b2 − 4ac = 0 → two equal real roots
(C) If b2 − 4ac < 0 → no real roots
(I) 2x2 −3x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain a = 2, b = −3, 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√3 x + 4 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain a = 3, b = −4√3, 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
(III) 2x2 − 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain 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 as follows.
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