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# Find the Values Of K For Which the Roots Are Real and Equal in Each of the Following Equation: (2k + 1)X2 + 2(K + 3)X + K + 5 = 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:

(2k + 1)x2 + 2(k + 3)x + (k + 5) = 0

#### Solution

The given quadric equation is (2k + 1)x2 + 2(k + 3)x + k + 5 = 0, and roots are real and equal

Then find the value of k.

Here,

a = (2k + 1), b = 2(k + 3) and c = k + 5

As we know that D = b2 - 4ac

Putting the value of a = (2k + 1), b = 2(k + 3) and c = k + 5

={2(k + 3)}2 - 4 x (2k + 1) x (k + 5)

= {4(k2 + 6k + 9)} - 4(2k2 + 11k + 5)

= 4k2 + 24k + 36 - 8k2 - 44k - 20

= -4k2 - 20k + 16

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

-4k2 - 20k + 16 = 0

-4(k2 + 5k - 4) = 0

k2 + 5k - 4 = 0

Now factorizing the above equation

k2 + 5k - 4

k=(-b+-sqrt(b^2-4ac))/(2a)

k=(-5+-sqrt(25+16))/2

k=-5+-sqrt41/2

So, either

Therefore, the value of k=-5+-sqrt41/2

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Solution Find the Values Of K For Which the Roots Are Real and Equal in Each of the Following Equation: (2k + 1)X2 + 2(K + 3)X + K + 5 = 0 Concept: Nature of Roots.
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