MHB *2 values of x for perpendicular vectors

[karush]

The vectors $$\left(\frac{2x}{x-3}\right)$$ and $$\left(\frac{x+1}{5}\right)$$ are $$\perp$$ for $$2$$ values of $$x $$

what is the quadratic equation which the 2 values of x satisfy

well I got $$2x^2+7x-15$$ which gives $$x=-5$$ and $$x=\frac{3}{2}$$
-5 seems to work but
$$\frac{3}{2}$$ doesn't or is there another way to do this.

thanks ahead,

 

[I like Serena]

Re: 2 values of x for perpendicular vectors

The vectors $$\left(\frac{2x}{x-3}\right)$$ and $$\left(\frac{x+1}{5}\right)$$ are $$\perp$$ for $$2$$ values of $$x $$

what is the quadratic equation which the 2 values of x satisfy

well I got $$2x^2+7x-15$$ which gives $$x=-5$$ and $$x=\frac{3}{2}$$
-5 seems to work but
$$\frac{3}{2}$$ doesn't or is there another way to do this.

thanks ahead,

Hi karush! :)

What is the reason you think $x=\frac 3 2$ does not work?
It is a proper solution.

 

[karush]

Re: 2 values of x for perpendicular vectors

when I plugged $$\frac{3}{2}$$ in the slopes didn't seem $$\perp$$

 

[I like Serena]

Re: 2 values of x for perpendicular vectors

when I plugged $$\frac{3}{2}$$ in the slopes didn't seem $$\perp$$

Well... which vectors do you get if you plug it in?

 

[karush]

Re: 2 values of x for perpendicular vectors

OK found my error...