My Math Forum Factor this second degree polynomial

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 August 26th, 2011, 07:00 AM #1 Newbie   Joined: Aug 2011 Posts: 3 Thanks: 0 Factor this second degree polynomial I hit an impasse when trying to isolate x from the following equation (dual second degree polynomial?): (1.5*x)^2+x^2=3.5^2 What rules do I apply, in order to factor (1.5*x)^2+x^2 into one square expression so that I may treat it as a classic second degree polynomial (ax^2 + bx + c = 0)? Using numberempire.com I am able to come up with concrete answers, the above resolves to 7/sqrt(13). However, I am a bit embarrassed not being able to remember my algebra and would love to know just how numberempire.com arrives at this answer. Could a kind math soul help out?
 August 26th, 2011, 07:08 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Factor this second degree polynomial [color=#000000]I don't see why it is so difficult, $\left(\frac{3}{2}x\right)^2+x^2=\left(\frac{7}{2}\ right)^{2}\Leftrightarrow \left(\frac{9}{4}+1\right)x^2=\left(\frac{7}{2}\ri ght)^{2}\Leftrightarrow \frac{13}{4}x^2=\frac{49}{4}\Leftrightarrow x^2=\frac{49}{13}\Leftrightarrow x=\pm \frac{7}{\sqrt{13}}$.[/color]
 August 26th, 2011, 07:23 AM #3 Newbie   Joined: Aug 2011 Posts: 3 Thanks: 0 Re: Factor this second degree polynomial Interesting... thanks ZardoZ! I was hoping to be able to rewrite it so I could plug it into a spreadsheet, letting 1.5 (a) and 3.5 (d) being variables... it doesn't look like that is easily done though: (a*x)^2+x^2=d^2 All I really want to arrive at, is calculating the area of a rectangle using only the ratio of the sides and the diagonal (as in, calculating the surface area of monitor, knowing only the physical resolution and diagonal size, say 1024x768 and 19"). Can this be done?
 August 26th, 2011, 10:12 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Factor this second degree polynomial You may solve for x: $(ax)^2+x^2=d^2$ $x^2$$a^2+1$$=d^2$ $x^2=\frac{d^2}{a^2+1}$ $x=\pm\frac{d}{\sqrt{a^2+1}}$ For finding the area of a rectangle armed only with the diagonal measurement d and the aspect ratio a (ratio of width x to height y), we may use: (1) $\frac{x}{y}=a$ (2) $x^2+y^2=d^2$ From (1), we find $x=ay$ and substituting into (2) we get: $(ay)^2+y^2=d^2$ $y^2=\frac{d^2}{a^2+1}$ Now, take x = ay, and multiply both sides by y: $xy=ay^2=\frac{ad^2}{a^2+1}$ Since the area A of the rectangle is xy, we have: $A=\frac{ad^2}{a^2+1}$
 August 26th, 2011, 10:29 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,931 Thanks: 1125 Math Focus: Elementary mathematics and beyond Re: Factor this second degree polynomial $\theta\,=\,\arctan$$\frac{V}{H}$$,\,V\text{ is vertical resolution},\,H\,\text{is horizontal resolution and }$ $D\,\text{is diagonal measurement}.$ $\text{Area}\,\approx\,(D\cos(\theta))(D\sin(\theta ))$ This should give a fair approximation of the view-able area of the monitor - horizontal scaling may be a little different than its vertical counterpart. The link below may help. http://en.wikipedia.org/wiki/Pixel_dens ... onitor_PPI
 August 30th, 2011, 11:36 AM #6 Newbie   Joined: Aug 2011 Posts: 3 Thanks: 0 Re: Factor this second degree polynomial Thanks a lot guys!

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