# Math Help - The quadratic Problem

I am not able to solve this problem

if $a^2+b^2=1$ then find the range of $a+b$

what I am doing is
$(a+b)^2>0$
$a^2+b^2+2ab>0$ where $a^2+b^2=1$
$1+2ab>0$
$2ab>-1$

Case 2
$(a-b)^2>0$
$a^2+b^2-2ab>0$ where $a^2+b^2=1$
$1-2ab>0$
$2ab<1$

$-1<2ab<1$

$a+b=\sqrt{a^2+b^2+2ab}$
$a+b=\sqrt{1+2ab}$
$a+b=\sqrt{1-1}$ $-1<2ab<1$ to find minimum
$a+b=0$

$a+b=\sqrt{1+1}$ $-1<2ab<1$ to find maximum
$a+b=\sqrt{2}$

The range I am getting is $(0,\sqrt{2})$ but the answer is ( $(-\sqrt{2},\sqrt{2})$

2. ## Re: The quadratic Problem

I would do this a completely different way. If we use x and y rather than a and b (it is still exactly the same problem, of course), we can think of "a^2+ b^2= 1" (x^2+ y^2= 1) as the equation of a circle with center at (0, 0) and radius 1. "a+ b= c" or x+ y= c, for c a fixed number, is a line sloping down to the left at a 45 degree angle. The smallest and largest possible values of c are when that line is tangent to the circle which happen at the endpoints of the diameter perpendicular to the line: that is the line y= x which crosses the circle $x^2+ y^2= 1$ when $x^2+ x^2= 2x^2= 1$ so that $x= y= \pm\sqrt{2}/2$. Taking $x= a= b= \sqrt{2}/2$, $a+ b= \sqrt{2}/2+\sqrt{2}/2= \sqrt{2}$. But taking $x= a= b= -\sqrt{2}/2$ we get $a+ b= -\sqrt{2}$.

Perhaps you forgot the $\pm$ on the square root?