# Quadratic Problem

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• July 30th 2013, 02:58 AM
varunkanpur
Quadratic Problem
I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.
• July 30th 2013, 08:23 AM
HallsofIvy
Re: Quadratic Problem
Are you aware that the graph of $x^2+ y^2= 1$ is a circle with center at (0, 0) and radius 1? So that the circle extends from x= -1 to x= 1 horizontally, and y= -1 to 1 vertically.
• July 30th 2013, 11:24 AM
topsquark
Re: Quadratic Problem
Quote:

Originally Posted by varunkanpur
I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.

Technical point: a and A are not the same variable.

-Dan
• July 30th 2013, 04:28 PM
Shakarri
Re: Quadratic Problem
Assuming you haven't done geometry of the circle you can think of it this way.

$A^2+B^2=1$

$A^2=1-B^2$

$A=\sqrt{1-B^2}$

Think of the maximum and minimum vales of B so that A is real. Then do the same thing for A instead of B.
• August 2nd 2013, 09:48 PM
smatik
Re: Quadratic Problem
hmm if you've done trigonometry,try this. Let A=sin x and B=cos x[Parametric equation of circle]. Now range of A+B is Range of sin x + cos x i.e [-1/sqrt(2),1/sqrt(2)].
• August 3rd 2013, 06:59 PM
johng
Re: Quadratic Problem
Hi,
smatak has a good idea, but unfortunately he gave the wrong answer.

$sin(x)+cos(x)=\sqrt2sin(x+\pi/4)$

So the range of sin(x)+cos(x) is $[-\sqrt2,\sqrt2]$
• August 3rd 2013, 11:37 PM
smatik
Re: Quadratic Problem
oh yea.i calculated wrong.... :( btw its smatik