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Find the Values Of K For Which the Given Quadratic Equation Has Real and Distinct Roots: Kx2 + 6x + 1 = 0 - Mathematics

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Question

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

kx2 + 6x + 1 = 0

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Solution

The given quadric equation is kx2 + 6x + 1 = 0, and roots are real and distinct.

Then find the value of k.

Here,

a = k, b = 6 and c = 1

As we know that D = b2 - 4ac

Putting the value of a = k, b = 6 and c = 1

D = (6)2 - 4 x (k) x (1)

= 36 - 4k

The given equation will have real and distinct roots, if D > 0

36 - 4k > 0

Now factorizing of the above equation

36 - 4k > 0

4k < 36

k < 36/4

k < 9

Now according to question, the value of k less than 9

Therefore, the value of k < 9.

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Chapter 4: Quadratic Equations - Exercise 4.6 [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercise 4.6 | Q 6.2 | Page 42

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