SOLUTION: Find k so that (k + 1)x^2 + 5x + 2k - 1 = 0 has two real solutions.

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Question 1124033: Find k so that (k + 1)x^2 + 5x + 2k - 1 = 0 has two real solutions.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find k so that+%28k+%2B+1%29x%5E2+%2B+5x+%2B+2k+-+1+=+0 has two+real solutions.
use discriminant:
if discriminant is negative,b%5E2+-+4ac%3C0, our equation has two complex solutions
if discriminant is positive,+b%5E2+-+4ac%3E+0 , our equation has two solutions which they are real
if discriminant is zero, b%5E2+-+4ac=+0 , our equation has one solution which is real


+%28k+%2B+1%29x%5E2+%2B+5x+%2B+2k+-+1+=+0...here, a=%28k+%2B+1%29,b=5, and c=2k+-+1+
b%5E2+-+4ac%3E0

5%5E2+-+4%2A%28k%2B1%29%282k+-+1%29%3E0
25+-+%284k%2B4%29%282k+-+1%29%3E0
25+-+%288k%5E2+-+4k%2B8k-4%29%3E0
25=8k%5E2+%2B4k-4
8k%5E2+%2B4k-4-25%3E0
8k%5E2+%2B4k-29%3E0
use quadratic formula:
k=%28-b%2B-sqrt%28b%5E2+-+4ac%29%29%2F2a
k=%28-4%2B-sqrt%284%5E2+-+4%2A8%28-29%29%29%29%2F%282%2A8%29
k=%28-4%2B-sqrt%2816+%2B32%2A29%29%29%2F16
k=%28-4%2B-sqrt%28944%29%29%2F16
k=%28-4%2B-30.72%29%2F16
solutions:
k=%28-4%2B30.72%29%2F16
k=26.72%2F16
k=1.67
k=%28-4-30.72%29%2F16
k=%28-34.72%29%2F16
k=-2.17
so, we will have b%5E2+-+4ac%3E0 if k=-2.17 and k=1.67


check:
+%28k+%2B+1%29x%5E2+%2B+5x+%2B+2k+-+1+=+0 ...if k=-2.17
+%28-2.17+%2B+1%29x%5E2+%2B+5x+%2B+2%28-2.17%29+-+1+=+0
+-1.17x%5E2+%2B+5x+-+4.34+-+1+=+0
+-1.17x%5E2+%2B+5x+-+5.34++=+0
solutions are:
x2.10
x2.18

graph:


+%28k+%2B+1%29x%5E2+%2B+5x+%2B+2k+-+1+=+0 ...if k=1.67
+%281.67+%2B+1%29x%5E2+%2B+5x+%2B+2%281.67%29+-+1+=+0
+2.67x%5E2+%2B+5x+%2B+3.34+-+1+=+0
+2.67x%5E2+%2B+5x+%2B+2.34++=+0
solutions:
x-0.95
x-0.92