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

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

Can anyone explain the answer, I. am unable to understand this question itself .Please help

Posted from my mobile device

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

Hi VeritasKarishma, Can you please provide a good and easy solution for the above question? Thanks.

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

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.

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

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

Posted from my mobile device

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

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

Now option D will make sense.

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

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!!

Solution:

Since 7 is a root of x^2 - 2px + m = 0, we have:

7^2 - 2p(7) + m = 0

49 - 14p + m = 0

49 + m = 14p

7 + m/7 = 2p

Since m is a multiple of 5 and p is a prime, we see that m must be a multiple of 7 also. In other words, m is a multiple of 35, However, since 2p is even and 7 is odd, m/7 must be odd (since odd + odd = even). In other words, m must be odd. Since m is less than 120, m can only be 35 x 1 = 35 or 35 x 3 = 105.

If m = 35, we have:

7 + 35/7 = 2p

12 = 2p

6 = p

However, since p is a prime, p can’t be 6, and hence m can’t be 35.

If m = 105, we have:

7 + 105/7 = 2p

22 = 2p

11 = p

So m must be 105, and p must be 11. Now, since 12 is a root of x^2 – 2px + n = 0 and we know p = 11, we have:

12^2 - 2(11)(12) + n = 0

144 - 264 + n = 0

-120 + n = 0

n = 120

Therefore, p + n - m = 11 + 120 - 105 = 26.

Answer: D

x^2-2px+m=0
Here, one root is given as 7. Let other root be a.
x^2-2px+m = (x-7)(x-a)
x^2-2px+m = x^2-(a+7)x+7a

Product of roots, m= 7*a
But m is divisible by 5. So, m= 7*5*b.

Given that m<120.
Therefore, 35b < 120 or b < 3.4 which means b is 1, 2 or 3.
If b=1, m=35
If b=2, m=70
If b=3, m=105

Sum of roots, -2px = -(a+7)
But we have a = 5b
This means p = (5b+7)/2
p is an integer, so b has to be odd.
If b=1, p=6 (Not possible since p is prime)
Therefore b=3, p=11 and m= 103.

Substitute x=12 and p=11 in x^2-2px+n=0,
n=120

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

x^2-2px+m=0, m=10,15....115
r1=7, =>m=35,70,105, r2=5,10,15
2p=12,17, 22
p=6(no),8.5(no), 11(yes)
m=105; p=11

144-22*12+n=0
n=120
p+n-m
11+120-105=11+15=26

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