Quadratic Problem

[varun_kanpur]

I am not able to solve this problem

if \(\displaystyle a^2+b^2=1\) then find the range of \(\displaystyle a+b\)

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

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

\(\displaystyle -1<2ab<1\)

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


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


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

 

[stapel]

if \(\displaystyle a^2+b^2=1\) then find the range of \(\displaystyle a+b\)

what I am doing is....

Try something simpler. What is the graph of x^2 + y^2 = 1? (Hint: Don't ignore the bottom half of the circle!)

 

[varun_kanpur]

Try something simpler. What is the graph of x^2 + y^2 = 1? (Hint: Don't ignore the bottom half of the circle!)

It is the equation of a circle with radius 1 with center at the origin.

 

[DrPhil]

It is the equation of a circle with radius 1 with center at the origin.

I was about to give the same hint, but I hadn't worked through the next step yet.

From symmetry, the max will either be where one of the squares is 1 and the other 0, OR where the two are equal. In terms of the circle, we are looking for the max of \(\displaystyle \sin\theta + \cos\theta\). That should lead directly to the answer.

The case with both a and b negative leads to the lower limit of the range.

 

[Deleted member 4993]

I am not able to solve this problem

if \(\displaystyle a^2+b^2=1\) then find the range of \(\displaystyle a+b\)

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

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

\(\displaystyle -1<2ab<1\)

\(\displaystyle a+b=\sqrt{a^2+b^2+2ab}\)
\(\displaystyle a+b=\sqrt{1+2ab}\)
\(\displaystyle a+b=\sqrt{1-1}\)
\(\displaystyle a+b=0\)


\(\displaystyle a+b \ = \ \) \(\displaystyle \pm\) \(\displaystyle \sqrt{1+1}\)
\(\displaystyle a+b \ = \ \) \(\displaystyle \pm\) \(\displaystyle \sqrt{2}\)


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

.