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# If X 2 + K ( 4 X + K − 1 ) + 2 = 0 Has Equal Roots, Then K = - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

If $x^2 + k\left( 4x + k - 1 \right) + 2 = 0$ has equal roots, then k =

• $- \frac{2}{3}, 1$

• $\frac{2}{3}, - 1$

• $\frac{3}{2}, \frac{1}{3}$

• $- \frac{3}{2}, - \frac{1}{3}$

#### Solution

The given quadric equation is $x^2 + k\left( 4x + k - 1 \right) + 2 = 0$, and roots are equal

Then find the value of k.

$x^2 + k\left( 4x + k - 1 \right) + 2 = 0$

x^2 + 4kx + (k^2 - k + 2) = 0

Here, a = 1 , b = 4k and , c = k^2 - k + 2

As we know that  D = b^2 - 4ac

Putting the value of  a = 1 , b = 4k and , c = k^2 - k + 2

=(4k)^2 - 4xx 1 xx (k^2 - k + 2)

= 16k^2 - 4k^2 + 4k - 8

=12k^2 + 4k - 8

=4 (3k^2 + k - 2)

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

4(3k^2 + k - 2) = 0

3k^2 + k - 2 = 0

3k^2 + 3k - 2k - 2 = 0

3k (k+1) - 2(k+ 1) = 0

(k + 1)(3k - 2) = 0

(k+ 1) = 0

k = -1

or

(3k - 2) = 0

k = 2/3

Therefore, the value of k =2/3: -1

Is there an error in this question or solution?

#### APPEARS IN

Solution If X 2 + K ( 4 X + K − 1 ) + 2 = 0 Has Equal Roots, Then K = Concept: Solutions of Quadratic Equations by Factorization.
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