# [SOLVED]Two values of x for perpendicular vectors

#### karush

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

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

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: 2 values of x for perpendicular vectors

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

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

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

Hi karush!

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

#### karush

##### Well-known member
Re: 2 values of x for perpendicular vectors

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: 2 values of x for perpendicular vectors

when I plugged $$\displaystyle \frac{3}{2}$$ in the slopes didn't seem $$\displaystyle \perp$$
Well... which vectors do you get if you plug it in?

#### karush

##### Well-known member
Re: 2 values of x for perpendicular vectors

OK found my error......