Question 925613
<pre>
That's one way of solving it.  Here's another.

Either way you'll have to check for extraneous solutions.

Let the two roots be x=a and x=b,

Then we have a system of three equations and 3 unknowns:

{{{system(a^2-12a-P=0,b^2-12b-P=0,a^2=b)}}}

Substitute {{{a^2}}} for {{{b}}} in

{{{b^2-12b-P=0}}}
{{{(a^2)^2-12(a^2)-P=0}}}
{{{a^4-12a^2-P=0}}}

Solve for P

{{{a^4-12a^2=P}}}

Substitute in

{{{a^2-12a-P=0}}}
{{{a^2-12a-(a^4-12a^2)=0}}}
{{{a^2-12a-a^4+12a^2=0}}}
{{{-a^4+13a^2-12a=0}}}
{{{a^4-13a^2+12a=0}}}
{{{a(a^3-13a+12)=0}}}

It's easy to see that 1 is a solution to {{{a^3-13a+12}}}

1 | 1  0 -13  12
  |<u>    1   1 -12</u>
    1  1 -12   0

So we have factored the polynomial equation as

{{{a(a-1)(a^2+a-12)=0}}}

We further factor it as

{{{a(a-1)(a+4)(a-3)=0}}}

So we have four solutions for a

a=0, a=1, a=-4, a=3

although one or more may be extraneous.

Substituting a=0 in

{{{a^4-12a^2=P}}}
{{{0^4-12(0)^2=P}}}
{{{0=P}}}

Substituting a=1 in

{{{a^4-12a^2=P}}}
{{{1^4-12(1)^2=P}}}
{{{1-12=P}}}
{{{-11=P}}}

Substituting a=-4 in

{{{a^4-12a^2=P}}}
{{{(-4)^4-12(-4)^2=P}}}
{{{256-12(16)=P}}}
{{{256-192=P}}}
{{{64=P}}}

Substituting a=3 in

{{{a^4-12a^2=P}}}
{{{3^4-12(3)^2=P}}}
{{{81-12(9)=P}}}
{{{81-108=P}}}
{{{-27=P}}}

------------------

Checking:
P=0
{{{x^2-12x-P=0}}}
{{{x^2-12x-0=0}}}
{{{x^2-12x=0}}}
{{{x(x-12)=0}}}
{{{x=0}}}, {{{x=12}}}

So P=0 is extraneous

P=-11
{{{x^2-12x-P=0}}}
{{{x^2-12x-(-11)=0}}}
{{{x^2-12x+11=0}}}
{{{(x-11)(x-1)=0}}}
{{{x=11}}}, {{{x=1}}}

So P=-11 is extraneous

P=64
{{{x^2-12x-P=0}}}
{{{x^2-12x-64=0}}}
{{{(x+4)(x-16)=0}}}
{{{x=-4}}}, {{{x=16}}}

P=64 is a solution, since (-4)<sup>2</sup>=16

P=-27
{{{x^2-12x-P=0}}}
{{{x^2-12x-(-27)=0}}}
{{{x^2-12x+27=0}}}
{{{(x-3)(x-9)=0}}}
{{{x=3}}}, {{{x=9}}}

P=-27 is a solution, since 3<sup>2</sup>=9

There are 2 actual solutions, P=64 and P=-27

Edwin</pre>