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Re: A quadratic equation is in the form [#permalink]
root of eq1: 7 and a.
roots of eq2: 12 and b.

7*a = m (Sum of roots formula)
7+a = 2p (Product of roots formula)

12*b = n (Sum of roots formula)
12+b = 2p (Product of roots formula)

Since m is divisible by 5 and less than 120. (m = 7*a)

a = 5,10,15

we also know that p is prime. p = (7+a)/2

substitute a= 5,10,15. and only a= 15 gives p as prime. p =11.
so a =15, b =10 (2p = 12+b)

m = 105 (m=7*a)
n = 120 (n= 12*b)
p = 11.

So, p+n-m = 26
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Re: A quadratic equation is in the form [#permalink]
2
dvk007 wrote:
A quadratic equation is in the form of \(x^2–2px + m = 0\), where m is divisible by 5 and is less than 120. One of the roots of this equation is 7. If p is a prime number and one of the roots of the equation, \(x^2–2px + n = 0\) is 12, then what is the value of \(p+n–m\)?

A. 0
B. 6
C. 16
D. 26
E. 27


For One-Answer-Multiple-Choice questions, we can take following steps.
1. Simplify conditions and the problem in the question first.
2. Try all possible cases one by one.
3. When we encounter a solution, we can choose it.


Since \(7\) is one root of the equation, \(x^2-2px+m=0\), we have \(49 - 14p + m = 0\) or \(m = 14p - 49\).
Since \(12\) is one root of the equation, \(x^2-2px+n=0\), we have \(144 - 24p + n = 0\) or \(n = 24p - 144\).
Thus we have \(n - m = (24p - 144) - (14p-49) = 10p -95\).

\(p + n - m = p + 10p - 95 = 11p - 95\).

When we look at choices, all of them are greater than or equal to \(0\).
Thus we have \(p \ge 9\), since \(p + n - m = 11p - 95 \ge 0\).
Now, we can try prime numbers greater than or equal to \(9\) one by one.
The first prime number greater than or equal to \(9\) is \(p = 11\) and we have \(11 \cdot p - 95 = 11 \cdot 11 - 95 = 121 - 95 = 26\).
\(26\) is one of the choices and we can choose D.

Therefore, D is the right answer.

Originally posted by AskMathQuestions on 24 Aug 2020, 15:02.
Last edited by AskMathQuestions on 25 Aug 2020, 07:52, edited 1 time in total.
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Re: A quadratic equation is in the form [#permalink]
GreenlightTestPrep wrote:
dvk007 wrote:
A quadratic equation is in the form of \(x^2–2px + m = 0\), where m is divisible by 5 and is less than 120. One of the roots of this equation is 7. If p is a prime number and one of the roots of the equation, \(x^2–2px + n = 0\) is 12, then what is the value of \(p+n–m\)?

A. 0
B. 6
C. 16
D. 26
E. 27


GIVEN: x = 7 is one of the roots of the equation x² – 2px + m = 0
This means (x - 7) must be one of the factors of the expression on the left side of the equation.
That is, x² – 2px + m = 0, can be rewritten as (x - 7)(x +/- something) = 0 [notice that x = 7 is definitely a solution to the new equation]
Let's assign the variable k to the missing number (aka "something")
We can write: x² – 2px + m = (x - 7)(x - k)

GIVEN: m is divisible by 5 and is less than 120
We already know that: x² – 2px + m = (x - 7)(x - k)
If we expand the right side we get: x² – 2px + m = x² – kx - 7x + 7k
Now rewrite the right side as follows: x² – 2px + m = x² – (k + 7)x + 7k

We can see that 2p = k + 7
And we can see that m = 7k

In order for m to be divisible by 5, it must be the case that k is divisible by 5.
So, k COULD equal 5, 10, 15, 20, 25, etc
Let's test a few possible values of k

If k = 5, then 2p = 5 + 7 = 12
When we solve this, we get: p = 6
HOWEVER, we're told that p is PRIME
So, it cannot be the case that k = 5

If k = 10, then 2p = 10 + 7 = 17
When we solve this, we get: p = 8.5
HOWEVER, we're told that p is PRIME
So, it cannot be the case that k = 10

If k = 15, then 2p = 15 + 7 = 22
When we solve this, we get: p = 11
Aha! 11 is PRIME
So, it COULD be the case that k = 15. Let's confirm that this satisfies the other conditions in the question.

If k = 15, then we get: x² – 2px + m = (x - 7)(x - 15)
Expand and simplify the right side: x² – 2px + m = x² – 22x + 105
So, this meets the condition that says m is divisible by 5 and is less than 120

We now know that p = 11 and m = 105
All we need to do now is determine the value of n

GIVEN: x = 12 is one of the solutions of the equation x² – 2px + n = 0
Plug in x = 12 to get: 12² – 2p(12) + n = 0
Since we already know that p = 11, we can replace p with 11 to get: 12² – 2(11)(12) + n = 0
Simplify: 144 - 264 + n = 0
Simplify: -120 + n = 0
Solve: n = 120

What is the value of p + n – m?
p + n – m = 11 + 120 - 105
= 26

Answer: D

Cheers,
Brent



Why did you assign "-k" as one of the roots to the equation. What can't it be "+k"?
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Re: A quadratic equation is in the form [#permalink]
I think there is a mistake in the two equations because both of the equations used (x). That could indicate that x in the first equation is the same x in the second equation. However, this is not the reality.

What do you think Carcass?
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Re: A quadratic equation is in the form [#permalink]
Expert Reply
No I do not think there is such possible confusion. It is beyond the scope of the question.

Frankly I do not think this is a GRE question. Not sure though.

Another simple approach could be

substitute x = 7 in first equation to get 49-14p+m = 0

Substitute x = 12 in second equation to get 144-24p+n = 0

n - m = 24p - 144 + 49 - 14p = 10p - 95

p + n - m = 11p - 95

p must be a prime number which when multiplied by 11 is greater than 95. prime values greater than 7 satisfy our requirement. p = 11 gets us option D.
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Re: A quadratic equation is in the form [#permalink]
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