Answer:
Please check the explanation!
Step-by-step explanation:
As the discrimination expression is
[tex]b^2-4ac[/tex]
if
[tex]b^2-4ac\:<0[/tex]
then the equation has no real solution.
Given the equation
[tex]mx^2+3x+1-m=0[/tex]
by comparing the quadratic equation
[tex]\:ax^2+bx+c=0[/tex]
[tex]mx^2+3x+1-m=0[/tex]
we can observe that
[tex]a=m[/tex]
[tex]b=3[/tex]
[tex]c=(1-m)[/tex]
substituting the values in discrimination to find the values of m
so
[tex]3^2-4m\left(1-m\right)=0[/tex]
[tex]9-4m+4m^2=0[/tex]
[tex]4m^2-4m+9=0[/tex]
[tex]\frac{-4m+4m^2}{4}=\frac{-9}{4}[/tex]
[tex]-m+m^2=-\frac{9}{4}[/tex]
[tex]\left(m-\frac{1}{2}\right)^2=-2[/tex]
as
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
so solving
[tex]m-\frac{1}{2}=\sqrt{-2}[/tex]
[tex]m-\frac{1}{2}=\sqrt{-1}\sqrt{2}[/tex]
[tex]m-\frac{1}{2}=\sqrt{2}i[/tex] ∵ [tex]\sqrt{-1}=i[/tex]
[tex]m=\sqrt{2}i+\frac{1}{2}[/tex]
and also solving
[tex]m-\frac{1}{2}=-\sqrt{-2}[/tex]
[tex]m-\frac{1}{2}=-\sqrt{2}i[/tex]
[tex]m=-\sqrt{2}i+\frac{1}{2}[/tex]
As we know that
if
[tex]b^2-4ac\:<0[/tex]
then the equation has no real solution.
Therefore, for the values of [tex]m=\sqrt{2}i+\frac{1}{2},\:m=-\sqrt{2}i+\frac{1}{2}[/tex], for which the equation [tex]mx^2+3x+1-m=0[/tex] will have no solution.
