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# Find the Values of K for Which the Quadratic Equation ( 3 K + 1 ) X 2 + 2 ( K + 1 ) X + 1 = 0 Has Equal Roots. Also, Find the Roots. - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

Find the values of k for which the quadratic equation

$\left( 3k + 1 \right) x^2 + 2\left( k + 1 \right)x + 1 = 0$ has equal roots. Also, find the roots.

#### Solution

The given quadric equation is  $\left( 3k + 1 \right) x^2 + 2\left( k + 1 \right)x + 1 = 0$ and roots are real and equal.

Then, find the value of k.

Here

$a = 3k + 1, b = 2(k + 1) \text { and } c = 1$.

As we know that

$D = b^2 - 4ac$

Putting the values of  $a = 3k + 1, b = 2(k + 1) \text { and } c = 1$

$D = \left[ 2\left( k + 1 \right) \right]^2 - 4\left( 3k + 1 \right)\left( 1 \right)$

$= 4( k^2 + 2k + 1) - 12k - 4$

$= 4 k^2 + 8k + 4 - 12k - 4$

$= 4 k^2 - 4k$

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

Thus,

$4 k^2 - 4k = 0$

$\Rightarrow 4k(k - 1) = 0$

$\Rightarrow k = 0 \text { or } k - 1 = 0$

$\Rightarrow k = 0 \text { or } k = 1$

Therefore, the value of k is 0 or 1.
Now, for k = 0, the equation becomes

$x^2 + 2x + 1 = 0$

$\Rightarrow x^2 + x + x + 1 = 0$

$\Rightarrow x(x + 1) + 1(x + 1) = 0$

$\Rightarrow (x + 1 )^2 = 0$

$\Rightarrow x = - 1, - 1$

for k = 1, the equation becomes

$4 x^2 + 4x + 1 = 0$

$\Rightarrow 4 x^2 + 2x + 2x + 1 = 0$

$\Rightarrow 2x(2x + 1) + 1(2x + 1) = 0$

$\Rightarrow (2x + 1 )^2 = 0$

$\Rightarrow x = - \frac{1}{2}, - \frac{1}{2}$

Hence, the roots of the equation are $- 1 \text { and } - \frac{1}{2}$.

Is there an error in this question or solution?

#### APPEARS IN

Solution Find the Values of K for Which the Quadratic Equation ( 3 K + 1 ) X 2 + 2 ( K + 1 ) X + 1 = 0 Has Equal Roots. Also, Find the Roots. Concept: Solutions of Quadratic Equations by Factorization.
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