SOLUTION: Given the function f(x) = 2x^2 + 3x - 1 , find the value(s) of k so that f(k+2) = -2. - I nee help how to start off and what lesson to use. Thank you!

Algebra ->  Systems-of-equations -> SOLUTION: Given the function f(x) = 2x^2 + 3x - 1 , find the value(s) of k so that f(k+2) = -2. - I nee help how to start off and what lesson to use. Thank you!      Log On


   

Question 943356: Given the function f(x) = 2x^2 + 3x - 1 , find the value(s) of k so that f(k+2) = -2.
- I nee help how to start off and what lesson to use. Thank you!
Found 2 solutions by reviewermath, MathLover1:
Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
Given the function f(x) = 2x^2 + 3x - 1 , find the value(s) of k so that f(k+2) = -2.
Solution:
In f(x) = 2x^2 + 3x - 1, replace x by (k+2), the result is
f(k+2) = 2(k+2)^2 + 3(k+2) - 1
-2 = 2(k^2 + 4k + 4) + (3k + 6) - 1
-2 = 2k^2 + 11k + 13
2k^2 + 11k + 15 = 0
(2k + 5)(k + 3) = 0
2k + 5 = 0 or k + 3 = 0
Therefore, k = -5/2 or k = -3

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Given the function f%28x%29+=+2x%5E2+%2B+3x+-+1 , find the value(s) of k so that f%28k%2B2%29+=+-2.
f%28k%2B2%29+=2%28k%2B2%29%5E2+%2B+3%28k%2B2%29+-+1=-2
2%28k%5E2%2B4k%2B4%29+%2B+3k%2B6-+1=-2
2k%5E2%2B8k%2B8+%2B+3k%2B6-+1=-2
2k%5E2%2B11k%2B13=-2
2k%5E2%2B11k%2B13%2B2=0
2k%5E2%2B11k%2B15=0
2k%5E2%2B6k%2B5k%2B15=0
%282k%5E2%2B6k%29%2B%285k%2B15%29=0
2k%28k%2B3%29%2B5%28k%2B3%29=0
%28k%2B3%29+%282k%2B5%29+=+0
solutions:
if %28k%2B3%29++=+0 =>+k=-3
if %282k%2B5%29+=+0 => k=-5%2F2

check one solution:
f%28k%2B2%29+=2%28k%2B2%29%5E2+%2B+3%28k%2B2%29+-+1 => k=-3
f%28-3%2B2%29+=2%28-3%2B2%29%5E2+%2B+3%28-3%2B2%29+-+1+
f%28-1%29+=2%28-1%29%5E2+%2B+3%28-1%29+-+1
f%28-1%29+=2+-3+-+1
f%28-1%29+=+-2+ ...as given
you can check for k=-5%2F2