Find the value (s) of k if the quadratic equation 3
- k 31+ 4 = 0 has real roots.
w
i
elefendine
em entende a right angle at the centre Find area of mind
Answers
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)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. \begin{lgathered}D=0\\b^2-4ac=0\end{lgathered}
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(2(k+1))
2
−4(3k+1)(1)=0
4(k^2+1+2k)-12k-4=04(k
2
+1+2k)−12k−4=0
4k^2+4+8k-12k-4=04k
2
+4+8k−12k−4=0
4k^2-4k=04k
2
−4k=0
4k(k-4)=04k(k−4)=0
k=0,4k=0,4
When k=0, The equation became x^2+2x+1=0x
2
+2x+1=0
The roots are
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}x=
2a
−b±
b
2
−4ac
x=\frac{-2\pm\sqrt{2^2-4(1)(1)}}{2(1)}x=
2(1)
−2±
2
2
−4(1)(1)
x=\frac{-2\pm\sqrt{0}}{2}x=
2
−2±
0
x=\frac{-2}{2}=-1x=
2
−2
=−1
Roots are -1,-1.
When k=4, The equation became 13x^2+10x+1=013x
2
+10x+1=0
The roots are
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}x=
2a
−b±
b
2
−4ac
x=\frac{-10\pm\sqrt{10^2-4(13)(1)}}{2(1)}x=
2(1)
−10±
10
2
−4(13)(1)
x=\frac{-10\pm\sqrt{48}}{26}x=
26
−10±
48
x=\frac{-10\pm4\sqrt{3}}{26}x=
26
−10±4
3
x=\frac{-10+4\sqrt{3}}{26},\frac{-10-4\sqrt{3}}{26}x=
26
−10+4
3
,
26
−10−4
3
x=\frac{-5+2\sqrt{3}}{13},\frac{-5-2\sqrt{3}}{13}x=
13
−5+2
3
,
13
−5−2
3
Roots are \frac{-5+2\sqrt{3}}{13},\frac{-5-2\sqrt{3}}{13}
13
−5+2
3
,
13
−5−2
3
not equal.
So, For k=0 roots are equal.