Question 1078806
<pre><b>
{{{x^2 + bx + (b/2)^2 = n}}}

{{{x^2 + bx + (b/2)^2 - n=0}}}

{{{x^2 + bx + ((b/2)^2 - n)=0}}} 

A quadratic equation with 0 on the right has two
real roots if the discriminant > 0

We find the discriminant which is {{{b^2-4ac}}}

here a=1, b=b, c={{{(b/2)^2 - n}}}

So the discriminant is

{{{b^2-4(1)((b/2)^2)^2}}} which we set {{{"">0}}}

{{{b^2-4(b/2)^4}}}{{{"">0}}}

{{{b^2-4(b^4/16)}}}{{{"">0}}}

{{{b^2-b^4/4}}}{{{"">0}}}

Multiply through by 4

{{{4b^2-b^4}}}{{{"">0}}}

Factor out b<sup>2</sup>

{{{b^2(4-b^2)}}}{{{"">0}}}

{{{b^2(2-b)(2+b)}}}{{{"">0}}}

The critical numbers are 0,2 and -2.

[The critical numbers all cause the discriminant to = 0,
which gives the quadratic only one real solution.
So none of the critical numbers are possible values 
for b.

We place the critical numbers on a number line

---------o-------o-------o--------
-4  -3  -2  -1   0   1   2   3   4

We test values in each interval:

Using test value -3 for the first (leftmost) interval

{{{b^2(2-b)(2+b)}}}{{{"">0}}}
{{{(-1)^2(2-(-1)^"")(2+(-1)^"")}}}{{{"">0}}}
{{{(1)(2+1)(2+1)}}}{{{"">0}}}
{{{(1)(3)(3)}}}{{{"">0}}}
{{{9}}}{{{"">0}}}

That's true so we shade the interval (-2,0).

---------o=======o-------o--------
-4  -3  -2  -1   0   1   2   3   4

Using test value 1 for the next interval

{{{b^2(2-b)(2+b)}}}{{{"">0}}}
{{{(1)^2(2-(1)^"")(2+(1)^"")}}}{{{"">0}}}
{{{(1)(2-1)(2+1)}}}{{{"">0}}}
{{{(9)(1)(3)}}}{{{"">0}}}
{{{27}}}{{{"">0}}}


That's true so we also shade the interval (0,2).

---------o=======o=======o--------
-4  -3  -2  -1   0   1   2   3   4


Using test value 3 for the next interval

{{{b^2(2-b)(2+b)}}}{{{"">0}}}
{{{(3)^2(2-(3)^"")(2+(3)^"")}}}{{{"">0}}}
{{{(9)(2-3)(2+3)}}}{{{"">0}}}
{{{(9)(-1)(5)}}}{{{"">0}}}
{{{-45}}}{{{"">0}}}

That's false.  Therefore b must be in these intervals
in order for the equation to have 2 real roots:
<font face="arial" size=6>
(-2,0) U (0,2)

Edwin</pre></font></b>