Question

find the value of k for which the quadratic equation (3k+1)x2 + 2 (k+1) x+1=0 has equal roots also find the roots

Answer 1

Answer:

Step-by-step explanation:below

Answer 2

Answer:

For k=0 roots are equal.        

Step-by-step explanation:

Given : Quadratic equation $(3k+1)x^2+2(k+1)x+1=0$ has equal roots.

To find : The value of k and roots?

Solution :

The quadratic equation has equal roots when discriminant is zero.

i.e. $D=0\\b^2-4ac=0$

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

$(2(k+1))^2-4(3k+1)(1)=0$

$4(k^2+1+2k)-12k-4=0$

$4k^2+4+8k-12k-4=0$

$4k^2-4k=0$

$4k(k-4)=0$

$k=0,4$

When k=0, The equation became $x^2+2x+1=0$

The roots are

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-2\pm\sqrt{2^2-4(1)(1)}}{2(1)}$

$x=\frac{-2\pm\sqrt{0}}{2}$

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

Roots are -1,-1.

When k=4, The equation became $13x^2+10x+1=0$

The roots are

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-10\pm\sqrt{10^2-4(13)(1)}}{2(1)}$

$x=\frac{-10\pm\sqrt{48}}{26}$

$x=\frac{-10\pm4\sqrt{3}}{26}$

$x=\frac{-10+4\sqrt{3}}{26},\frac{-10-4\sqrt{3}}{26}$

$x=\frac{-5+2\sqrt{3}}{13},\frac{-5-2\sqrt{3}}{13}$

Roots are $\frac{-5+2\sqrt{3}}{13},\frac{-5-2\sqrt{3}}{13}$ not equal.

So, For k=0 roots are equal.