Question

pls solve this quadratic equation...
Find the value of k when equation has equal roots...
x²+4kx+(k²-k+2)=0

Answer 1

Here is your answer.

a=1, b=4K

C=k^2-k+2

b^2-4ac=0

After substituting we get,

16k^2-4k^2-4K +8

After evaluating we get an equation:

12k^2-4K+8. (On diving with 4)

3k^2 -k + 2 =0

Then on evaluation we get value of k

-1 or 2/3

Answer 2

☆EQUATION :-

-------------------

${x}^{2} + 4kx + ( {k }^{2} - k + 2) = 0$

From above equation.

A = 1 , B = 4k , C= k^2 - k +2

Now,

°.° It has equal roots .

.°. B^2 - 4AC =0

Equating the values.

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

=> 16k^2 - 4k^2 +4k -8 =0

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

=>4 [3k^2 + k -2 ] =0

=> 3k^2 +k -2 =0

=> 3k^2 +3k -2k -2 = 0

=> 3k (k +1) -2 (k+1) =0

=> (3k-2)(k+1) =0

=> 3k-2=0 OR k+1 = 0

=> k = 2/3 OR k = -1

So, the value of 'k' is 2/3 0r -1.

****HOPE YOU UNDERSTAND ****

THANK YOU