Question

find the value of k for quadratic equation kx^+(2k+4)×+9=0 so that they have two equal roots​

Answer 1

Answer:

kx²+(2k+4)+9=0

kx²+2k+4+9=0

kx²+2k+13=0

kx²+2k= -13

kx²= -13-2k

x²=-13-2k/k

x²=-13-2

x²=-15

x=√-15

Answer 2

k can have two values 1 and 4 for the equation to have equal roots.

Step-by-step explanation:

We are given the following quadratic equation in the question:

$kx^2 +(2k+4)x+9=0$

General form of quadratic equation:

$ax^2 + bx + c = 0\\a = k\\b = 2k + 4\\c = 9$

If the quadratic equation have two equal roots then, the discriminant of the root is zero.

Equating discriminant to zero, we get,

$D = b^2 - 4ac\\(2k+4)^2 - 4(k)(9) = 0\\4k^2 + 16 +16k-36k = 0\\4k^2-20k +16 = 0\\k^2 - 5k + 4 - 0\\k(k-4)-1(k-4) = 0\\(k-4)(k-1) = 0\\k = 1, 4$

Thus, k can have two values 1 and 4 for the equation to have equal roots.

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