Answer:
Please check the explanation
Step-by-step explanation:
Given the expression
[tex]\frac{x^2}{x-3}=\frac{x+2}{2x-5}[/tex]
[tex]\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c[/tex]
[tex]x^2\left(2x-5\right)=\left(x-3\right)\left(x+2\right)[/tex]
[tex]2x^3-5x^2=x^2-x-6[/tex]
[tex]2x^3-6x^2+x+6=0[/tex]
[tex]\left(x-2\right)\left(2x^2-2x-3\right)=0[/tex]
Using the zero factor principle:
if [tex]ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
[tex]x-2=0\quad \mathrm{or}\quad \:2x^2-2x-3=0[/tex]
so
[tex]x-2=0[/tex]
[tex]x=2[/tex]
and
[tex]2x^2-2x-3=0\:\\x=\frac{1+\sqrt{7}}{2},\:x=\frac{1-\sqrt{7}}{2}[/tex]
so
[tex]x=2,\:x=\frac{1+\sqrt{7}}{2},\:x=\frac{1-\sqrt{7}}{2}[/tex]
