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Question 151895: [p+q][square]=10
p[square]+q[square]=50
Find p and q .
PLS HELP T.T
Thankx: [p+q][square]=10
p[square]+q[square]=50
Find p and q .
PLS HELP T.T
Thankx
Answer by Edwin McCravy(2036) About Me  (Show Source):
You can put this solution on YOUR website!
system((p+q)^2=10, p^2+q^2=50)
Find p and q .
PLS HELP T.T
Thankx

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