Discussion

If p and q are positive, p2 + q2 = 16, and p2q2 = 8, then q =

(A)   2

(B)...
(C)...
(D)...
(E)...
(F)...
*This question is included in Introduction to Nova GRE Math, question #10

The solution is

Posted: 06/24/2012 21:46
Can you explain how you are subtracting the equations from one another? I am just not seeing it from the example given.
Posted: 06/25/2012 02:33
Step by step:
Posted: 06/25/2012 09:45
thank you so much
Posted: 07/24/2012 15:18
Why do you subtract the equations?
Posted: 07/24/2012 16:50
Katie, recall from Algebra. If you have two equations with two unknowns, you can eliminate one unknown at a time in order to solve for the remaining unknown.
Posted: 11/05/2012 21:17
Can you explain why you are dividing the equation by 2? Why 2? Why divide at all?
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Contributor
Posted: 11/05/2012 22:22
Mike, we divide the equation by 2 because we want to solve for q, so divide by 2 to get rid of q's unit variable.
Posted: 11/24/2012 09:41
When you subtract the equation how do you get 2q squared. One q is positive and the other is negative... Shouldn't they cancel each other out?
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Contributor
Posted: 11/24/2012 17:51
Janice,
Common properties:
4 + 4 = 8
4 - 4 = 0
4 + (+4) = 8
4 - (-4) = 8
4 + (-4) = 0
4 - (+4) = 0
Thus addition will cancel out q
p^2 + q^2 = 16
p^2 - q^2 = 8
--------------------- addition
p^2 + (+p^2) = 2p^2
q^2 + (-q^2) = 0
16 + (+8) = 24

Niels
Posted: 12/03/2012 13:37
Can we apply substitution as well?
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Contributor
Posted: 12/03/2012 15:20
Ngoc,
p^2 + q^2 = 16
p^2 = 16 - q^2
(16 - q^2) - q^2 = 8
16 - q^2 - q^2 = 8
-2q^2 = 8 - 16
-2q^2 = -8
q^2 = 4
q = sqrt(4)
q = 2

Niels
Posted: 03/19/2013 19:31
What mistake am I making wrong here:

I set the equations equal to q^2.
Q^2=16-p^2
Q^2=p^2-8

16-p^2=p^2-8
24=2p^2
12=p^2

Why doesn't this work?
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Contributor
Posted: 03/19/2013 19:35
Gia, you didn't do anything wrong. The question is asking for q, not p^2. So you just need to continue a few more steps.
Posted: 07/03/2013 07:59
Is the substitution method inapplicable here cuz I tried it and I got q=1....I need help
Posted: 07/03/2013 08:02
It's clear to me now
Posted: 09/25/2013 07:54
Is this how the question is going to be stated in the test? How do I know to subtract the second equation from the first?
Posted: 10/07/2013 16:44
Cara, this is one type of question that will be asked in the test. This is basic algebra, solving for a system of equation with unknowns. When you have two equations with two unknowns (a and b, for example), you can solve the equations for a and b. There are 2 ways to solve: 1) by subtracting one from another to eliminate a or b; 2) by substitution.
Posted: 10/20/2014 05:17
Does this solution violate order of operations by dividing both sides by 2 before taking the square route of both sides? I don't see any parenthesis or am I missing something?
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Contributor
Posted: 10/20/2014 12:30
Brandon, in the second step from last, the explanation divides both sides of the equation by 2, in order to get q^2=4. Then you factor q^2-4=0, to get (q-2)(q+2)=0, and q=-2 and 2.

I do not like the wording of the explanation (taking the square root of both sides), because mathematically speaking it is not accurate. The square root of 4 is 2, not -2 and 2. But when we factor out q^2-4=0, we do get q=-2 and q=2.

I hope that helps.
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Admin
Posted: 10/21/2014 01:01
Hi Brandon,

The order of operations acronym PEMDAS is the general rule for performing operations in an expression.

When solving equations, the rule is performed in reverse order: SADMEP.

So, we divide before taking the square root. However, taking the square root first would give the same answer. The guidelines SADMEP and PEMDAS can often be violated.

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