Real numbers m, p, and q are such that m + p = -q

Question

Real numbers  m ,  p , and  q  are such that  m+p=-q , and the sum of  mp ,  mq , and  pq  is  -60 .

What is the arithmetic mean of  m^2 ,  p^2 , and  q^2 ?

A.  20 
B.  40 
C.  60 
D.  120 
E.  180

onlyPlanA · Answer

Official explanation please

kevincan · Answer

You probably know  (x + y)^2 = x^2 + y^2 + 2xy

What about   (x + y + z)^ 2  ?

egmat · Answer

Hi freebunny,

Happy to walk through the full path. The whole problem turns on  one identity , and once you spot it the rest is just plugging in.

Step 1 - Rewrite the first condition

You're told `m + p = −q`. Move the `q` over and it becomes:

- `m + p + q = 0`

That clean "equals zero" is the key that unlocks everything.

Step 2 - Square both sides

Squaring a sum of three terms follows the rule:

- `(m + p + q)2 = m2 + p2 + q2 + 2(mp + mq + pq)`

Since the left side is `02 = 0`, we get:

- `m2 + p2 + q2 + 2(mp + mq + pq) = 0`

Step 3 - Use the second condition

You're given that `mp + mq + pq = −60`. Substitute it in:

- `m2 + p2 + q2 + 2(−60) = 0`
- `m2 + p2 + q2 − 120 = 0`
- `m2 + p2 + q2 =  120 `

Step 4 - Take the mean

The arithmetic mean of the three squares is their sum divided by  3 :

- ` 120  /  3  =  40 `

So the answer is  B .

Watch the last step  -  120  is sitting right there as choice  D . That's the trap:  120  is the  sum  of the squares, not the  mean . The question asks for the average, so you still have to divide by 3.

Quick way to lock in the identity:  try expanding `(a + b + c)2` on your own and confirm you get `a2 + b2 + c2 + 2(ab + ac + bc)`. Recognizing that "sum of squares + 2·(sum of pairwise products)" pattern is what makes a problem like this fast.

Answer: B

dhart · Answer

Yes, it loos like an important formaula in this case

chetan2u · Answer

Almost always, there is more than one way to solve a question and I always make it a point to practice them with my students.

Here too a very simple solution is as below...

Let q = 0, so m+p = -q = 0, that is m = -p.
Now mp+mq+pq =-60 
Putting q as 0, we get mp=-60
This tells one of the two, m and p, is  \sqrt{60}  and other - \sqrt{60}

We are to find \frac{m^2+p^2+q^2}{3}= ( \sqrt{60}^2 + (-\sqrt{60})^2+0)/3

=(60+60)/3 =40

B

undeniablevirgin · Answer

We're given:
 
 m + p + q = 0 
 mp + mq + pq = −60  
 Step 1: Express each variable in terms of the other two's sum 
Since m + p + q = 0, we get three relations:
 
 m + p = −q 
 m + q = −p 
 p + q = −m  
 Step 2: Substitute into mp + mq + pq = −60, grouping by each variable 
Group as m(p + q) + pq:
 
 m(−m) + pq = −60 
 pq − m2 = −60 
  pq + 60 = m2  ... (1)  
Group as q(m + p) + mp:
 
 q(−q) + mp = −60 
 mp − q2 = −60 
  q2 = mp + 60  ... (2)  
Similarly, grouping as p(m + q) + mq gives:
 
  p2 = mq + 60  ... (3)  
 Step 3: Add equations (1), (2), and (3) 
=> pq + 60 + mp + 60 + mq + 60 = m2 + p2 + q2
 => (mp + mq + pq) + 180 = m2 + p2 + q2
 => -60 + 180 = m2 + p2 + q2
 => m2 + p2 + q2 = 120

Step 4: Find the arithmetic mean 
Mean = (m2 + p2 + q2) / 3 = 120 / 3 = 40

Answer: B

Real numbers m, p, and q are such that m + p = -q

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Real numbers m, p, and q are such that m + p = -q

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