Question

find for what values of K does the quadratic equation X square + 2 X + K square minus 3 equal to zero has real and equal roots​

Answer 1

Answer:

Step-by-step explanation:

As the question says, the roots of the equation are equal. Let us take the roots as\alpha, \beta

Now we can say that \alpha=\beta=m

Therefore, the equation is x^{2}+k 2 x+k^{2}-k+2 which means if put according to a x^{2}+b x+c,

Then the sum of root is, \alpha+\beta=-\frac{b}{a}b

that is 2 m={2 k}{1} and

alpha x beta={c}{a}={k^{2}-k+2}{1}.

Therefore,  

Put the value of m = k^{2}=k^{2}-k+2 we get

k^{2}=k^{2}-k+2

k=2  

-----------------------------MARK ME AS BRAINY-------------------------------

Therefore, the value of k = 2.

Answer 2

Answer:K = 2

Step-by-step explanation:

*X^2+2X+K^2-3=0*

Compare with the general form

A=1 B=2 C=K^2 -3

For the roots are REAL and EQUAL

B^2-4AC=0

(2)^2 -4 (1) (K^2-3) = 0

4 - 4K^2 + 12 =0

-4K^2 +16 = 0

-4K^2 = -16

K^2 = -16/-4

K^2 = 4

sqrt(K) = sqrt(4) .....Taking square

root

K= 2