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

Algebra.Com
Question 1124033: Find k so that (k + 1)x^2 + 5x + 2k - 1 = 0 has two real solutions.
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
Find so that has solutions.
use discriminant:
if discriminant is negative,, our equation has two complex solutions
if discriminant is positive, , our equation has solutions which they are real
if discriminant is zero, , our equation has one solution which is real


...here, ,, and








use quadratic formula:





solutions:






so, we will have if and


check:
...if



solutions are:



graph:


...if



solutions:






RELATED QUESTIONS

Find the values of k so that the equation {{{(k-2)x^2+4x-2k+1=0}}} has two distinct real... (answered by josgarithmetic)
The equation x^2+(k+2)x+2k=0 has two distinct real roots. Find the possible values of... (answered by ikleyn)
Find the value(s) of k such that the equation f(x + k) = 0 has two equal real roots.... (answered by stanbon)
Find the value of k so that the line through the given points has slope m. (k+1,... (answered by stanbon)
Find k so that the roots of (2k-1)x^2 + (k-1)x - 16 = 0 are numerically equal but... (answered by ikleyn)
Find the value of k, given 4x - y +2k +1 =0 has real roots (answered by josgarithmetic)
Find the value(s) of k so that x^2 + 4(k-1)x + 4k^2 = 0 has no real... (answered by Fombitz)
find the value of k that leads to two equal real roots for... (answered by josgarithmetic)
show that x^2+(3k-2)x+k(k-1)=0 has real roots for all value of... (answered by fcabanski)