Question 1181072
Which of the following equation has greatest and smallest number of real solutions?

(a) {{{x^3 + x - 10 = 0}}}...factor

{{{x^3 +2x^2-2x^2+5x-4x - 10 = 0}}}

{{{(x^3 -2x^2)+(2x^2-4x) +(5x- 10) = 0}}}

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

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

one real solution is:{{{x=2}}}

find two more solutions for {{{(x^2 + 2x + 5) = 0}}}

use discriminant {{{b^2-4ac}}}, if {{{ b^2-4ac>0}}} there will be  real solutions

in your case {{{a=1}}}, {{{b=2}}}, {{{c=5}}}
{{{ 2^2-4*1(5)>0}}}
{{{ 4-20>0}}}
{{{ -16>0}}}-> {{{false}}}, so other two solutions are complex solutions



(b) {{{x^2 + 4x - 15 = 0}}}

use discriminant {{{b^2-4ac}}}, if {{{ b^2-4ac>0}}} there will be  real solutions

in your case {{{a=1}}}, {{{b=4}}}, {{{c=-15}}}

{{{ 4^2-4*1(-15)>0}}}
{{{ 16+60>0}}}
{{{ 76>0}}}->true, so there are {{{highlight(2)}}} real solutions

(c) {{{e^x - x = 0}}}
{{{e^x =x }}}....take natural log of both sides
{{{ln(e^x) =ln(x )}}}
{{{x*ln(e) =ln(x )}}}
{{{x*1 =ln(x )}}}
{{{x=ln(x )}}}->{{{highlight(No)}}} solution for {{{x}}} in {{{R}}}

(d) {{{sec(x) - e^(-x^2)}}}

set equal to zero
{{{sec(x) - e^(-x^2)=0}}}
{{{sec(x) =e^(-x^2)}}}
only if {{{x=0}}}-> one real solution


answer: 
the  equation b) {{{x^2 + 4x - 15 = 0}}} has greatest  number of real solutions
the  equation c) {{{e^x - x = 0}}} has smallest  number of real solutions