the value of k if kx2 + 2(k+1)x + (3k+1)= 0 has real and equal roots is
Answers
Answer: Quadratic Equations
Step-by-step explanation:
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Answer:
For k=0 roots are equal.
Step-by-step explanation:
Given : Quadratic equation (3k+1)x^2+2(k+1)x+1=0(3k+1)x2+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. $$\begin{lgathered}D=0\\b^2-4ac=0\end{lgathered}$$
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