Q. 195.0( 5 Votes )

Find the values of k for which the given quadratic equation has real and distinct roots:

Answer :

(i) Given: kx2 + 6x + 1 = 0

Comparing with standard quadratic equation ax2 + bx + c = 0


a = k b = 6 c = 1


For real and distinct roots: D > 0


Discriminant D = b2 – 4ac > 0


62 – 4k > 0


36 – 4k > 0


4k < 36


k < 9


(ii) Given: x2 – kx + 9 = 0


Comparing with standard quadratic equation ax2 + bx + c = 0


a = 1 b = – k c = 9


For real and distinct roots: D > 0


Discriminant D = b2 – 4ac > 0


(– k)2 – 4.1.9 = k2 – 36 > 0


k2 > 36


k > 6or k < – 6 taking square root both sides


(iii) 9x2 + 3kx + 4 = 0


Comparing with standard quadratic equation ax2 + bx + c = 0


a = 9 b = 3k c = 4


For real and distinct roots: D > 0


Discriminant D = b2 – 4ac > 0


(3k)2 – 4.4.9 = 9k2 – 144 > 0


9k2 > 144


k2 > 16


k > 4ork < – 4 taking square root both sides


(iv) 5x2 – kx + 1 = 0


Comparing with standard quadratic equation ax2 + bx + c = 0


a = 5 b = – k c = 1


For real and distinct roots: D > 0


Discriminant D = b2 – 4ac > 0


(– k)2 – 4.5.1 = k2 – 20 > 0


k2 > 20


k > 2√5 or k < –2√5 taking square root both sides


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