Question

find the value of K for which the following equation has equal roots . x²+4 Kx+(K²-K+2)=0​

Answer 1

Answer:

$k=\frac{2}{3} (or)-1$

Step-by-step explanation:

The given equation is :   $x^{2} +4kx+(k^{2}-k+2)=0$

Comparing this quadratic equation with $ax^{2}+bx +c=0$ , we get

=> $a = 1;b=4k;c=k^{2}-k+2$

For the equation to have equal roots, the discriminant,$D=b^{2}-4ac$ should be equal to zero(0).

So, $b^{2}-4ac=0 => b^{2}=4ac$

=> $(4k)^{2}=4(1)(k^{2}-k+2)$

=> $16k^{2}=4k^{2}-4k+8$

=> $12k^{2}+4k-8=0$

=> $3k^{2}+k-2=0$

=> $k = \frac{-1+\sqrt{1+24} }{2*3}(or) \frac{-1-\sqrt{1+24} }{2*3}$

=> $k = \frac{-1+5}{6} (or)\frac{-1-5}{6}$

=> $k=\frac{2}{3} (or)-1$

Answer 2

Step-by-step explanation:

The given equation is : x^{2} +4kx+(k^{2}-k+2)=0x

2

+4kx+(k

2

−k+2)=0

Comparing this quadratic equation with ax^{2}+bx +c=0ax

2

+bx+c=0 , we get

=> a = 1;b=4k;c=k^{2}-k+2a=1;b=4k;c=k

2

−k+2

For the equation to have equal roots, the discriminant,D=b^{2}-4acD=b

2

−4ac should be equal to zero(0).

So, b^{2}-4ac=0 => b^{2}=4acb

2

−4ac=0=>b

2

=4ac

=> (4k)^{2}=4(1)(k^{2}-k+2)(4k)

2

=4(1)(k

2

−k+2)

=> 16k^{2}=4k^{2}-4k+816k

2

=4k

2

−4k+8

=> 12k^{2}+4k-8=012k

2

+4k−8=0

=> 3k^{2}+k-2=03k

2

+k−2=0

=> k = \frac{-1+\sqrt{1+24} }{2*3}(or) \frac{-1-\sqrt{1+24} }{2*3}k=

2∗3

−1+

1+24

(or)

2∗3

−1−

1+24

=> k = \frac{-1+5}{6} (or)\frac{-1-5}{6}k=

6

−1+5

(or)

6

−1−5

=> k=\frac{2}{3} (or)-1k=

3

2

(or)−1