Answer:
[tex]x=\frac{8}{3}\ cm[/tex]
Step-by-step explanation:
we know that
The area of the right triangle ABC is equal to
[tex]A=\frac{1}{2}(AB)(BC)[/tex]
we have
[tex]A=17\ cm^2[/tex]
[tex]AB=(3x-2)\ cm[/tex]
[tex]BC=(x+3)\ cm[/tex]
substitute the values
[tex]17=\frac{1}{2}(3x-2)(x+3)[/tex]
[tex]34=(3x-2)(x+3)[/tex]
[tex]34=3x^2+9x-2x-6[/tex]
[tex]3x^2+7x-6-34=0[/tex]
[tex]3x^2+7x-40=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^2+7x-40=0[/tex]
so
[tex]a=3\\b=7\\c=-40[/tex]
substitute in the formula
[tex]x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}[/tex]
[tex]x=\frac{-7(+/-)\sqrt{529}} {6}[/tex]
[tex]x=\frac{-7(+/-)23} {6}[/tex]
[tex]x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}[/tex]
[tex]x=\frac{-7(-)23} {6}=-5[/tex]
therefore
The solution is
[tex]x=\frac{8}{3}\ cm[/tex]
