who can solve it........?

Question

(x+1)(x+2)=k(2x+7)

Answer 1

What are you solving for?
x? or k?
k is easy to solve for. Assuming that x does not equal -7/2, then
k = (x+1)(x+2)/(2x+7)
Solving for x requires a bit more work.
First FOIL the left side and distribute the k on the right side.
x^2 + 3x + 2 = 2kx + 7k
Now move everything to the left side.
x^2 + (3-2k)x + (2-7k) = 0
You can now use the quadratic formula.
a = 1, b = 3-2k and c = 2-7k
x = [-(3-2k) +/- sqrt((3-2k)^2 - 4(1)(2-7k))]/(2*1)
= [ 2k - 3 +/- sqrt(9 - 12k + 4k^2 - 8 + 28k)]/2
= [2k - 3 +/- sqrt(4k^2 +16k + 1)]/2

Answer 2

(x + 1)(x + 2) = k(2x + 7)

solve for x (I assume)

x^2 + 3x + 2 = 2kx + 7k

x^2 + (3 - 2k)x + (2 - 7k) = 0

(x + 1 1/2 - k)^2 + (2 - 7k) - (2 1/4 + k^2 - 3k) = 0

(x + 1 1/2 - k)^2 = k^2 + 4k + 1/4
[right hand side cannot easily be factored]

x = -1 1/2 + k +/- sqrt(k^2 + 4k + 1/4)

Answer 3

(x + 1)(x + 2) = k(2x + 7)
x^2 + 2x + x + 2 = (2k)x + 7k
x^2 + 3x + 2 = (2k)x + 7k
x^2 + 3x + 2 - (2k)x - 7k = 0
x^2 + (3 - (2k))x + (2 - 7k) = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(3 - (2k)) ± sqrt((3 - (2k))^2 - 4(1)(2 - 7k)))/(2(1))
x = (-(3 - 2k) ± sqrt(((3 - 2k)(3 - 2k)) - 4(2 - 7k))/2
x = (-(3 - 2k) ± sqrt((9 - 6k - 6k + 4k^2) - 8 + 28k))/2
x = (-(3 - 2k) ± sqrt((9 - 12k + 4k^2 - 8 - 28k))/2
x = (-(3 - 2k) ± sqrt(4k^2 - 40k + 1))/2


x = (-3 + 2k ± sqrt((4k^2 - 40k + 1)))/2
or
k = ((x + 1)(x + 2))/(2x + 7)

Answer 4

is k a constant or variable?
if k is a variable you shall need at least one more equation to solve for x and k.

Answer 5

i ran out of room on my post-it but i'm pretty sure you use direct variation...solve for k; and then solve for x.

Answer 6

What are you solving for X or K ?

Answer 7

im in grade 2 and u expect me to answer this?

Answer 8

x= [-(3-2k) +/-sqrt of ((3-2k)^2-4(7k+2)]/2.

Answer 9

i dont know...im late for school...