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if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is .....a.2 b.-2 c 1/2 d.none ​

User Mintz
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1 Answer

4 votes

Answer:

k=2

Problem:

if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is ..

Explanation:

Since the coefficient of x^2 is 1, we can use this identity to aid us: x^2+bx+(b/2)^2=(x+b/2)^2.

So we want the following:

[(k+2)/2]^2=2k

Apply the power on the left:

(k+2)^2/4=2k

Multiply both sides by 4:

(k+2)^2=8k

Expand left side:

k^2+4k+4=8k *I used identity (x+c)^2=x^2+2xc+c^2

Subtract 8k on both sides:

k^2-4k+4=0

Factor using the identity mentioned a couple lines above:

(k-2)^2=0

Since zero squared is zero, we want k-2=0.

Adding both sides by 2 gives k=2.

User Jesher
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7.4k points
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