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

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

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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
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 31 May 2019, 09:44
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7*x2 =m then x2 can be 5,10 or 15 so m can be 35, 70 or 105.

Next, 7+X2 =2p, where p is prime. P is prime when x2 =15. Then p =11, m=105.

Another equation:
xsqroot - 22x +n =0. Given x1=12
Then 12+x2 =22
x2 =10
x1 * x2 = n
n = 12*10 = 120

Now we know all the values and can determine p+n-m = 11+120-105=26.

D

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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 02 Jun 2019, 02:50
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LidiiaShchichko wrote:
7*x2 =m then x2 can be 5,10 or 15 so m can be 35, 70 or 105.

Next, 7+X2 =2p, where p is prime. P is prime when x2 =15. Then p =11, m=105.

Another equation:
xsqroot - 22x +n =0. Given x1=12
Then 12+x2 =22
x2 =10
x1 * x2 = n
n = 12*10 = 120

Now we know all the values and can determine p+n-m = 11+120-105=26.

D

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7*x2 =m then x2 can be 5,10 or 15 so m can be 35, 70 or 105.

Next, 7+X2 =2p, where p is prime. P is prime when x2 =15. Then p =11, m=105.

Another equation:
xsqroot - 22x +n =0. Given x1=12
Then 12+x2 =22
x2 =10
x1 * x2 = n
n = 12*10 = 120

Now we know all the values and can determine p+n-m = 11+120-105=26.

Could you explain how to come up with the equations marked in red? I don't really understand the solution
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 02 Jun 2019, 11:06
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x^2–2px + m=0
Products of the root=m
let the second roots of above equation is 'a'.
7*a=m
Because m is a multiple of 5, a must be multiple of 5 too.
a=5k , where k is an integer

7*5k<120
35k<120
k<3.xyz


Also 2p is the sum of both roots and p is a prime number.
1. As 7+5k is an integer, 5k must also be an integer.
2. k can only take value 3 because then sum of roots, 2p is 22 and value of p is 11, which is a prime number.

m= 7*15=105

x^2–2px + n = 0
one root is 12, and let second root is b
12+b=22
b=10
n=12*10=120

p+n-m=11+120-105=26

kiran120680 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
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 06 Jun 2019, 03:28
Can anyone explain the answer, I. am unable to understand this question itself .Please help

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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 06 Jun 2019, 10:28
3
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 *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 et al )

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

Ans D
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 08 Jun 2019, 03:24
kiran120680 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



Hi VeritasKarishma, Can you please provide a good and easy solution for the above question? Thanks.
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 08 Jun 2019, 08:17
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kiran120680 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
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 08 Jun 2019, 12:02
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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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A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 08 Jun 2019, 21:04
GMATPrepNow wrote:
kiran120680 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






Hello Brent,

I got that approach.But, I was trying to find out the gap in my reasoning.


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

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

You can multiple first equation with -1 and then add it to second equation .

This gives

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

Now adding 'p' on both sides gives

p+n-m = 95 - 9p

Now p =7 gives p+n-m as 27 and not 26 which is option D.


Please help.

Thanks
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 09 Jun 2019, 01:36
If we consider equation one and two as q1 and q2, and subtract them n-m =0 (x^2-2px +m -x^2+2px -n=0) thus the value of p+n-m should be p which is a prime number and none of the options are prime number kindly explain it

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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 09 Jun 2019, 06:34
prabsahi wrote:
GMATPrepNow wrote:
kiran120680 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






Hello Brent,

I got that approach.But, I was trying to find out the gap in my reasoning.


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

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

You can multiple first equation with -1 and then add it to second equation .

This gives

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

Now adding 'p' on both sides gives

p+n-m = 95 - 9p

Now p =7 gives p+n-m as 27 and not 26 which is option D.


Please help.

Thanks




n - m = 10p - 95. not 95 - 10p.

so p + n - m = 11p - 95.

Now option D will make sense.
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 10 Jun 2019, 05:55
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kiran120680 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


x^2 - 2px + m = 0 has root 7. Plug in 7 in this equation.
49 - 14p + m = 0 ... (I)
m = 14p - 49
Now note that m is a multiple of 5. So 14p - 49 is a multiple of 5.
14p will be even so it should have 4 as the units digit to get m as a multiple of 5. Hence p should have 1 as the units digit so first such prime number will be 11. If p = 11, we get m = 105 (less than 120). Any other value of p will give m > 120.

x^2 - 2px + n = 0 has root 12. Plug in 12 in this equation.
144 - 24*11 + n = 0 ... (II)
n = 120

p + n - m = 11 + 120 - 105 = 26

Answer (D)
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 10 Jun 2019, 06:16
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VeritasKarishma wrote:
kiran120680 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


x^2 - 2px + m = 0 has root 7. Plug in 7 in this equation.
49 - 14p + m = 0 ... (I)
m = 14p - 49
Now note that m is a multiple of 5. So 14p - 49 is a multiple of 5.
14p will be even so it should have 4 as the units digit to get m as a multiple of 5. Hence p should have 1 as the units digit so first such prime number will be 11. If p = 11, we get m = 105 (less than 120). Any other value of p will give m > 120.

x^2 - 2px + n = 0 has root 12. Plug in 12 in this equation.
144 - 24*11 + n = 0 ... (II)
n = 120

p + n - m = 11 + 120 - 105 = 26

Answer (D)



Thanks for the great explanation.

I got that approach.But, I was trying to find out the gap in my reasoning.


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

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

You can multiple first equation with -1 and then add it to second equation .

This gives

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

Now adding 'p' on both sides gives

p+n-m = 95 - 9p

Now p =7 gives p+n-m as 27 and not 26 which is option D.


Please help.

Thanks
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 10 Jun 2019, 07:07
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prabsahi wrote:
VeritasKarishma wrote:
kiran120680 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


x^2 - 2px + m = 0 has root 7. Plug in 7 in this equation.
49 - 14p + m = 0 ... (I)
m = 14p - 49
Now note that m is a multiple of 5. So 14p - 49 is a multiple of 5.
14p will be even so it should have 4 as the units digit to get m as a multiple of 5. Hence p should have 1 as the units digit so first such prime number will be 11. If p = 11, we get m = 105 (less than 120). Any other value of p will give m > 120.

x^2 - 2px + n = 0 has root 12. Plug in 12 in this equation.
144 - 24*11 + n = 0 ... (II)
n = 120

p + n - m = 11 + 120 - 105 = 26

Answer (D)



Thanks for the great explanation.

I got that approach.But, I was trying to find out the gap in my reasoning.


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

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

You can multiple first equation with -1 and then add it to second equation .

This gives

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

Now adding 'p' on both sides gives

p+n-m = 95 - 9p

Now p =7 gives p+n-m as 27 and not 26 which is option D.


Please help.

Thanks


if 49-14p+m = 0 and 144-24p+n = 0, then how are you getting n-m=-24p+14p +144-49= 95 - 10p?

n = 24p-144 and m = 14p-49 , which makes n-m = 24p-144-14p+49 = 10p-95.

p+n-m = 11p-95
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is  [#permalink]

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New post 10 Jun 2019, 07:12
Thanks[/quote]

if 49-14p+m = 0 and 144-24p+n = 0, then how are you getting n-m=-24p+14p +144-49= 95 - 10p?

n = 24p-144 and m = 14p-49 , which makes n-m = 24p-144-14p+49 = 10p-95.

p+n-m = 11p-95[/quote]


I realise my mistake.Its extremely silly and pointless.

Thanks a lot for highlighting!!

I just wrote back the equations and it was all good.Thanks again!!
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Re: A quadratic equation is in the form of x^2–2px + m = 0, where m is   [#permalink] 10 Jun 2019, 07:12
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