Question 1136708
Consider the quadratic equation {{{ax^2+10x+(a-5)}}}

Find the values for {{{a}}} for which there are
 
1) two solutions
2) no solutions (no real solution)
3) 1 solution (a repeated real number solution)

use discriminant: {{{b^2 - 4ac}}}

if discriminant >{{{ 0}}} then {{{2}}} solutions
if discriminant = {{{0 }}}then {{{1 }}}solution
if discriminant < {{{0 }}}then no real solution

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

in your case {{{ax^2+10x+(a-5)}}} , {{{a=a}}}, {{{b=10}}}, {{{c=(a-5)}}}

{{{10^2 - 4a(a-5) > 0}}}
{{{100 - 4a^2-20a > 0}}}
{{{100  > 4a^2-20}}}
{{{4a^2+20a-100<0}}}
{{{a^2+5a-25<0}}}.....use quadratic formula to get

=>{{{(1/2) (5 - 5 sqrt(5))<a<(1/2) (5 + 5 sqrt(5))}}}

=>{{{-3.09<a<8.09}}}

check the graph using {{{a=-3}}}

{{{-3x^2+10x+(-3-5)}}}

={{{-3x^2+10x-8}}}


{{{ graph( 600, 600, -6, 5, -10, 10, -3x^2+10x-8) }}}

so, there are two solutions



{{{10^2 - 4a(a-5) = 0}}}
{{{100 - 4a^2-20a = 0}}}
{{{100  = 4a^2-20}}}
{{{4a^2+20a-100=0}}}
{{{a^2+5a-25=0}}}.....use quadratic formula to get

=>{{{a=(1/2) (5 - 5 sqrt(5))}}} or {{{a=(1/2) (5 + 5 sqrt(5))}}}

=>{{{a=-3.09}}} or {{{a=8.09}}}

check the graph using {{{a=-3.09}}}

{{{-3.09x^2+10x+(-3.09-5)}}}

={{{-3.09x^2+10x-8.09}}}


{{{ graph( 600, 600, -6, 5, -10, 10, -3.09x^2+10x-8.09) }}}



so, there is one solution 


{{{10^2 - 4a(a-5) < 0}}}
{{{100 - 4a^2-20a <0}}}
{{{100  < 4a^2-20}}}
{{{4a^2+20a-100>0}}}
{{{a^2+5a-25>0}}}.....use quadratic formula to get

=>{{{a>(1/2) (5 +5 sqrt(5))}}} or {{{a<(1/2) (5 - 5 sqrt(5))}}}

=>{{{a>8.09}}} or {{{a>-3.09}}}

check the graph using {{{a=9}}}

{{{9x^2+10x+(9-5)}}}

={{{9x^2+10x+4}}}


{{{ graph( 600, 600, -6, 5, -10, 10, 9x^2+10x+4) }}}



so, there are no real solutions