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[SOLVED] Two values of x for perpendicular vectors

karush

Well-known member
Jan 31, 2012
2,771
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.

thanks ahead,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
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.

thanks ahead,
Hi karush! :)

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

karush

Well-known member
Jan 31, 2012
2,771
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
Mar 5, 2012
8,866
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
Jan 31, 2012
2,771
Re: 2 values of x for perpendicular vectors

OK found my error......