A quadratic equation is in the form of x^2 – 2px + m = 0, where m

[BTGModeratorVI]

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

Answer: D
Source: E-gmat

[Brent@GMATPrepNow]

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

Answer: D
Source: E-gmat

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

[swerve]

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

Answer: D
Source: E-gmat

\(x^2–2px + m=0\)

\(m\) is divisible by \(5\) and less than \(120\) so \(m\) max can be \(115.7\) is one of the roots of the equation so,

product of roots \(= m\)
\(7 \cdot a=m\)

So \(m\) can be \(35,70,115\)
and \(a\) can be \(5, 10,15\)

Sum of the roots
\(7 + a = 2p\)

In this case \(7 + 5\) and \(7+10\) doesn't make \(p\) as prime but \(7 +15\) does \(22 = 2p\)
and hence \(p\) is \(11\)


Or also we can do it like

\(m\) can be \(35,70\) and \(115\)

Also \(7^2–2p7 + m=0\)
\(49 - 14p +m=0\)
\(49 +m =14p\)
\(p\) is a prime (\(2,3,5,7,11\ldots\) )

\(49 + 105 = 14p\)
\(p=11\)

\(m=105\)

Second equation \(x^2–2px + n = 0\)
one root is \(12\)
so
\(12^2–2p*12 + n = 0\)
\(144 -24*11 +n =0\)
\(144-264+n= 0\)
\(120=n\)

so
\(p +n -m\)
\(=11+120 -105\)
\(=26\)

Therefore, D