Question 1124033
 Find {{{k}}} so that{{{ (k + 1)x^2 + 5x + 2k - 1 = 0}}} has {{{two}}}{{{ real}}} solutions.

use  discriminant:

if  discriminant is negative,{{{b^2 - 4ac<0}}}, our equation has two complex solutions

 if discriminant is positive,{{{ b^2 - 4ac> 0}}} , our equation has {{{two}}} solutions which they are real 

 if discriminant is zero, {{{b^2 - 4ac= 0}}} , our equation has one solution which is real 



{{{ (k + 1)x^2 + 5x + 2k - 1 = 0}}}...here, {{{a=(k + 1)}}},{{{b=5}}}, and {{{c=2k - 1 }}}

 {{{b^2 - 4ac>0}}}


{{{5^2 - 4*(k+1)(2k - 1)>0}}}

{{{25 - (4k+4)(2k - 1)>0}}}

{{{25 - (8k^2 - 4k+8k-4)>0}}}

{{{25=8k^2 +4k-4}}}

{{{8k^2 +4k-4-25>0}}}

{{{8k^2 +4k-29>0}}}

use quadratic formula:

{{{k=(-b+-sqrt(b^2 - 4ac))/2a}}}

{{{k=(-4+-sqrt(4^2 - 4*8(-29)))/(2*8)}}}

{{{k=(-4+-sqrt(16 +32*29))/16}}}

{{{k=(-4+-sqrt(944))/16}}}

{{{k=(-4+-30.72)/16}}}

solutions:

{{{k=(-4+30.72)/16}}}
{{{k=26.72/16}}}
{{{k=1.67}}}

{{{k=(-4-30.72)/16}}}
{{{k=(-34.72)/16}}}

{{{k=-2.17}}}

so, we will have {{{b^2 - 4ac>0}}} if  {{{k=-2.17}}} and {{{k=1.67}}}



check:

{{{ (k + 1)x^2 + 5x + 2k - 1 = 0}}} ...if {{{k=-2.17}}}

{{{ (-2.17 + 1)x^2 + 5x + 2(-2.17) - 1 = 0}}}

{{{ -1.17x^2 + 5x - 4.34 - 1 = 0}}}

{{{ -1.17x^2 + 5x - 5.34  = 0}}}

solutions are:

{{{x}}}&#8776;{{{2.10}}}
{{{x}}}&#8776;{{{2.18}}}


graph:

{{{drawing( 600,600, -1,3, -1, 1,
circle(2.1,0,.03),circle(2.18,0,.03),
 graph( 600,600, -1,3, -1, 1, -1.17x^2 + 5x - 5.34) ) }}}


 {{{ (k + 1)x^2 + 5x + 2k - 1 = 0}}} ...if {{{k=1.67}}}

 {{{ (1.67 + 1)x^2 + 5x + 2(1.67) - 1 = 0}}}

{{{ 2.67x^2 + 5x + 3.34 - 1 = 0}}}

{{{ 2.67x^2 + 5x + 2.34  = 0}}}

solutions:

{{{x}}}&#8776;{{{-0.95}}}
{{{x}}}&#8776;{{{-0.92}}}


{{{drawing( 600,600, -2,1, -1, 1,
circle(-0.95,0,.02),circle(-0.92,0,.02),
 graph( 600,600, -2,1, -1, 1, 2.67x^2 + 5x + 2.34) ) }}}