Answer:
b. option Is the correct answer right
This is college? Woah I'm not advanced. Middle school!
Answer:
k=2
Problem:
if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is ..
Step-by-step 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.
Answer:
b = c(a - 6x )
Step-by-step explanation:
Given
6x + 
= a ( subtract 6x from both sides )
= a - 6x ( multiply both sides by c to clear the fraction )
b = c(a - 6x)
