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Find the Values Of K For Which the Roots Are Real and Equal in Each of the Following Equation: (K + 1)X2 - 2(3k + 1)X + 8k + 1 = 0 - CBSE Class 10 - Mathematics

Question

Find the values of k for which the roots are real and equal in each of the following equation:

(k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0

Solution

The given quadric equation is (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0, and roots are real and equal

Then find the value of k.

Here,

a = k + 1, b = -2(3k + 1)x and c = 8k + 1

As we know that D = b2 - 4ac

Putting the value of a = k + 1, b = -2(3k + 1)x and c = 8k + 1

= (-2(3k + 1))2 - 4 x (k + 1) x (8k + 1)

= 4(9k2 + 6k + 1) - 4(8k2 + 9k + 1)

= 36k2 + 24k + 4 - 32k2 - 36k - 4

= 4k2 - 12k

The given equation will have real and equal roots, if D = 0

4k2 - 12k = 0

k2 - 3k = 0

Now factorizing of the above equation

k(k - 3) = 0

So, either

k = 0

Or

k - 3 = 0

k = 3

Therefore, the value of k = 0, 3.

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Solution Find the Values Of K For Which the Roots Are Real and Equal in Each of the Following Equation: (K + 1)X2 - 2(3k + 1)X + 8k + 1 = 0 Concept: Nature of Roots.
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