Question 155037
{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=k^2-4(2)(9)}}} Plug in {{{a=2}}}, {{{b=k}}} and {{{c=9}}}



{{{D=k^2-8(9)}}} Multiply 4 and 2 to get 8



{{{D=k^2-72}}} Multiply 8 and 9 to get 72



So we'll use the equation {{{D=k^2-72}}} to solve the following:


1) "find the values of k for which the quadratic {{{2x^2+kx+9=0}}} has one real solution"



If a quadratic has one real solution, then the discriminant is equal to zero. So this means that {{{D=0}}}



{{{D=k^2-72}}} Start with the previous equation



{{{0=k^2-72}}} Plug in {{{D=0}}}



{{{72=k^2}}} Add 72 to both sides



{{{0+-sqrt(72)=k}}} Take the square root of both sides



{{{k=sqrt(72)}}} or {{{k=-sqrt(72)}}} Break up the "plus/minus"



{{{k=2*sqrt(18)}}} or {{{k=-2*sqrt(18)}}} Simplify the square root



So if {{{k=2*sqrt(18)}}} or {{{k=-2*sqrt(18)}}}, then the discriminant is equal to zero. This means that the equation {{{2x^2+2*sqrt(18)x+9=0}}} or {{{2x^2-2*sqrt(18)x+9=0}}} only has one real solution.



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2) "find the values of k for which the quadratic {{{2x^2+kx+9=0}}} no real solutions"



If a quadratic has no real solutions, this means that the discriminant is less than zero. In other words, {{{D<0}}}



{{{D=k^2-72}}} Go back to the previous equation



{{{k^2-72<0}}} Since  {{{D<0}}}, this means that the right side is less than zero


{{{k^2<72}}} Add 72 to both sides



{{{k<0+-sqrt(72)}}} Take the square root of both sides



{{{k<sqrt(72)}}} or {{{k>-sqrt(72)}}} Break up the "plus/minus"



{{{k<2*sqrt(18)}}} or {{{k>-2*sqrt(18)}}} Simplify the square root



{{{-2*sqrt(18)<k<2*sqrt(18)}}} Recombine the two inequalities to form one compound inequality


So if "k" is in between {{{-2*sqrt(18)}}} and {{{2*sqrt(18)}}}, then the discriminant is less than zero. This means that if "k" is in between {{{-2*sqrt(18)}}} and {{{2*sqrt(18)}}}, then the quadratic {{{2x^2+kx+9=0}}} will have no solutions