Question 151895
{{{system((p+q)^2=10, p^2+q^2=50)}}} 
Find p and q . 
PLS HELP T.T 
Thankx
<pre><font size = 4 color = "indigo"><b>
{{{system((p+q)^2=10,p^2+q^2=50)}}}

Rewrite the left side of the first equation as {{{(p+q)(p+q)}}}:

{{{(p+q)(p+q)=10}}}

Use FOIL:

{{{p^2+pq+pq+q^2=10}}}

Combine middle terms, which are like terms:

{{{p^2+2pq+q^2=10}}}

So now the system is:

{{{system(p^2+2pq+q^2=10, p^2+q^2=50)}}} 

Multiply the second equation through by {{{-1}}}

{{{system(p^2+2pq+q^2=10,-p^2-q^2=-50)}}}

Add the left and right sides of the system:

{{{2pq=-40}}}

Solve for {{{q}}} by dividing both sides by {{{2p}}}:

{{{2pq/(2p)=-40/(2p)}}}

Simplifying:

{{{q=-20/p}}}

Substituting {{{-20/p}}} for q in

{{{p^2+q^2=50}}}

{{{p^2+(-20/p)^2=50}}}

{{{p^2 + (-20)^2/p^2 = 50}}}

{{{p^2 + 400/p^2 = 50}}}

Multiply through every term by {{{LCD=p^2}}}

{{{p^4 + 400 = 50p^2}}}

Get 0 on the right by adding {{{-50p^2}}}
to both sides:

{{{p^4-50p^2+400=0}}}

The left side factors as

{{{(p^2-10)(p^2-40)=0}}}

Setting each factor = 0:

{{{P^2-10=0}}}
{{{p^2=10}}}
{{{p}}}=±{{{sqrt(10)}}}

{{{P^2-40=0}}}
{{{p^2=40}}}
{{{p}}}=±{{{sqrt(40)}}}
{{{p}}}=±{{{sqrt(4*10)}}}
{{{p}}}=±{{{sqrt(4)sqrt(10)}}}
{{{p}}}=±{{{2sqrt(10)}}}

So the four values for {{{p}}} are

{{{p=sqrt(10)}}}, {{{p=-sqrt(10)}}}, {{{p=2sqrt(10)}}}, {{{p=-2sqrt(10)}}}

We must now find a value of q to go with
each of these.  So we use

{{{q=-20/p}}}

Substituting {{{p=sqrt(10)}}} 

{{{q=-20/sqrt(10)}}}

Rationalize the denominator by multiplying top and
bottom by {{{sqrt(10)}}}

{{{q=-20sqrt(10)/sqrt(10)sqrt(10)}}}

{{{q=-20sqrt(10)/10}}}

{{{q=-2sqrt(10)}}}

So one solution is (p,q) = ({{{sqrt(10)}}},{{{-2sqrt(10)}}})

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

Substituting {{{p=-sqrt(10)}}} 

{{{q=-20/(-sqrt(10))}}}

Rationalize the denominator by multiplying top and
bottom by {{{sqrt(10)}}}

{{{q=-20sqrt(10)/(-sqrt(10)sqrt(10))}}}

{{{q=-20sqrt(10)/(-10)}}}

{{{q=2sqrt(10)}}}

So a second solution is (p,q) = ({{{-sqrt(10)}}},{{{2sqrt(10)}}})

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

Substituting {{{p=2sqrt(10)}}} 

{{{q=-20/(2sqrt(10))}}}

{{{q=-10/sqrt(10)}}}

Rationalize the denominator by multiplying top and
bottom by {{{sqrt(10)}}}

{{{q=-10sqrt(10)/(sqrt(10)sqrt(10))}}}

{{{q=-10sqrt(10)/10}}}

{{{q=-sqrt(10)}}}

So a third solution is (p,q) = ({{{2sqrt(10)}}},{{{-sqrt(10)}}})

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

Substituting {{{p=-2sqrt(10)}}} 

{{{q=-20/(-2sqrt(10))}}}

{{{q=10/sqrt(10)}}}

Rationalize the denominator by multiplying top and
bottom by {{{sqrt(10)}}}

{{{q=10sqrt(10)/(sqrt(10)sqrt(10))}}}

{{{q=10sqrt(10)/10}}}

{{{q=sqrt(10)}}}

So the fourth solution is (p,q) = ({{{-2sqrt(10)}}},{{{sqrt(10)}}})

Edwin</pre>