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Bunuel
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 20 May 2020, 07:06
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D
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The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15


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D
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vp1989
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 19 Jul 2020, 00:48
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Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15
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we know that
a∗b=12,
a+b=p,
|a−b|≥12,
upon squaring above equation
\(a^2+b^2−2ab≥144\)
\(a^2+b^2≥168\).........(i)

\((a+b)^2=a^2+b^2+2ab\)
Using inequality from eqn (i) above
\((a+b)^2=a^2+b^2+2ab≥168+2ab\)
\((a+b)^2≥192\)
i.e \(p^2≥192\)
values of p can be 14,15,16.....or−14,−15,−16
Now, since p is a negative number, the max. value of p to satisfy this will be−14.

So Option D.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x Updated on: 06 Jun 2020, 23:56
Originally posted by SiffyB on 20 May 2020, 08:28.
Last edited by SiffyB on 06 Jun 2020, 23:56, edited 1 time in total.
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Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15

Are You Up For the Challenge: 700 Level Questions
Show more

We know that,
sum of roots = a + b = p, product of roots = a * b = 12

Now, find out the values of a and b which satisfy the equation a * b = 12
They are:
(2, 6), (3, 4), (1, 12), (-2, -6), (-3, -4), (-1, -12)
Since, p is negative, a + b has to be negative as well.
We will focus only on the negative values of the roots a and b

p = a + b

From the values of the roots above, we have p = -7, -8 and -13
The greatest integer of these three is -7
IMO, (A)
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 21 May 2020, 19:35
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Given: \(x^2−px+12=0\) assume p = -q, since p is a negative constant.
\(x^2+qx+12 = 0,\) roots of the equation are
\(x_{1} = a = \frac{-q+\sqrt{(q^2-48)}}{2}, x_{2} = b = \frac{-q-\sqrt{(q^2-48)}}{2}\)
Given |a−b|≥12
\((a-b) = [\frac{-q+\sqrt{(q^2-48)}}{2}] -[ \frac{-q-\sqrt{(q^2-48)}}{2}] = \sqrt{(q^2-48)}\)
\(|(a-b)| = |\sqrt{(q^2-48)}|,\) since the quantity under square root should be positive, we will look for value of q, for which \(\sqrt{(q^2-48)}\geq{12} \)
\(for, q = 7 ---> \sqrt{(q^2-48)} = 1 \)
\(for, q = 8 ---> \sqrt{(q^2-48)} = 4 \)
\(for, q = 13 --> \sqrt{(q^2-48)} = 11 \)
\(for, q = 14 --> \sqrt{(q^2-48)} = \sqrt{148} \geq{12} ---> satisfies the condition \)
\(for, q = 15 --> \sqrt{(q^2-48)} = \sqrt{177} \geq{12} ---> satisfies the condition \)
Now, p = -q, so, for p = -14 and -15, the condition is satisfied, but we need greatest possible integer, so between -14 & -15, greatest possible integer is -14.
So, IMO, answer is D
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 19 Jul 2020, 03:59
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kris19
Given: \(x^2−px+12=0\) assume p = -q, since p is a negative constant.
\(x^2+qx+12 = 0,\) roots of the equation are
\(x_{1} = a = \frac{-q+\sqrt{(q^2-48)}}{2}, x_{2} = b = \frac{-q-\sqrt{(q^2-48)}}{2}\)
Given |a−b|≥12
\((a-b) = [\frac{-q+\sqrt{(q^2-48)}}{2}] -[ \frac{-q-\sqrt{(q^2-48)}}{2}] = \sqrt{(q^2-48)}\)
\(|(a-b)| = |\sqrt{(q^2-48)}|,\) since the quantity under square root should be positive, we will look for value of q, for which \(\sqrt{(q^2-48)}\geq{12} \)
\(for, q = 7 ---> \sqrt{(q^2-48)} = 1 \)
\(for, q = 8 ---> \sqrt{(q^2-48)} = 4 \)
\(for, q = 13 --> \sqrt{(q^2-48)} = 11 \)
\(for, q = 14 --> \sqrt{(q^2-48)} = \sqrt{148} \geq{12} ---> satisfies the condition \)
\(for, q = 15 --> \sqrt{(q^2-48)} = \sqrt{177} \geq{12} ---> satisfies the condition \)
Now, p = -q, so, for p = -14 and -15, the condition is satisfied, but we need greatest possible integer, so between -14 & -15, greatest possible integer is -14.
So, IMO, answer is D
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I don't know if I'm missing something here but how'd you eliminate the 2 in the denominator when subtracting a from b. Shouldn't the 2 still remain there?
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 19 Jul 2020, 04:12
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Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15


Are You Up For the Challenge: 700 Level Questions
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Please Bunuel can you help explain this and share the best way to solve this?
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 19 Jul 2020, 04:15
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NoobyMcScooby
kris19
Given: \(x^2−px+12=0\) assume p = -q, since p is a negative constant.
\(x^2+qx+12 = 0,\) roots of the equation are
\(x_{1} = a = \frac{-q+\sqrt{(q^2-48)}}{2}, x_{2} = b = \frac{-q-\sqrt{(q^2-48)}}{2}\)
Given |a−b|≥12
\((a-b) = [\frac{-q+\sqrt{(q^2-48)}}{2}] -[ \frac{-q-\sqrt{(q^2-48)}}{2}] = \sqrt{(q^2-48)}\)
\(|(a-b)| = |\sqrt{(q^2-48)}|,\) since the quantity under square root should be positive, we will look for value of q, for which \(\sqrt{(q^2-48)}\geq{12} \)
\(for, q = 7 ---> \sqrt{(q^2-48)} = 1 \)
\(for, q = 8 ---> \sqrt{(q^2-48)} = 4 \)
\(for, q = 13 --> \sqrt{(q^2-48)} = 11 \)
\(for, q = 14 --> \sqrt{(q^2-48)} = \sqrt{148} \geq{12} ---> satisfies the condition \)
\(for, q = 15 --> \sqrt{(q^2-48)} = \sqrt{177} \geq{12} ---> satisfies the condition \)
Now, p = -q, so, for p = -14 and -15, the condition is satisfied, but we need greatest possible integer, so between -14 & -15, greatest possible integer is -14.
So, IMO, answer is D

I don't know if I'm missing something here but how'd you eliminate the 2 in the denominator when subtracting a from b. Shouldn't the 2 still remain there?
Show more

Hi NoobyMcScooby,

I believe you are referring to this step:
\((a-b) = [\frac{-q+\sqrt{(q^2-48)}}{2}] -[ \frac{-q-\sqrt{(q^2-48)}}{2}] = \sqrt{(q^2-48)}\)

Here, the two -q will be subtracted and will go from the equation. However, the two \(\frac{\sqrt{(q^2-48)}}{2}\) will be added together. When that happens, we get a 2 in the numerator which will get canceled with the 2 in denominator.

Hope that helps!
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 19 Jul 2020, 04:22
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Yup, that was the mistake I was making. Thank you so much man.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 26 Aug 2020, 01:21
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SiffyB
Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15

Are You Up For the Challenge: 700 Level Questions

We know that,
sum of roots = a + b = p, product of roots = a * b = 12

Now, find out the values of a and b which satisfy the equation a * b = 12
They are:
(2, 6), (3, 4), (1, 12), (-2, -6), (-3, -4), (-1, -12)
Since, p is negative, a + b has to be negative as well.
We will focus only on the negative values of the roots a and b

p = a + b

From the values of the roots above, we have p = -7, -8 and -13
The greatest integer of these three is -7
IMO, (A)
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How did you know a and b were integers? And none of your ordered pairs listed has a distance greater than or equal to 12.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 03 Jun 2021, 10:55
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so this one of those questions where the devil is in the details.

firstly p is an integer, and then we are looking for the maximum value of p. now remember that -1 > -2, so your actually looking for a negative value which is as close to 0. now you also know that p is an integer so the most obviously solution seems to be C as x^2-13x-12=0 will have roots of -1 and -12.

BUT! -12 and -1 have a difference of 11, and you need something where the difference is at least 12 |a-b|>=12

so if you've identified this you're only left with D and E, and you could have guessed if you're stuck.

now you need to start putting the quadratics to work. a*b = 12 and p=a+b. so (a-b)^2>=12^2,
do the substitutions and you'll end up with p^2>=168 an again you need a negative value which is the close to 0, hence -13.

Pretty tricky question.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 15 Jun 2021, 01:44
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SiffyB
Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15

Are You Up For the Challenge: 700 Level Questions

We know that,
sum of roots = a + b = p, product of roots = a * b = 12

Now, find out the values of a and b which satisfy the equation a * b = 12
They are:
(2, 6), (3, 4), (1, 12), (-2, -6), (-3, -4), (-1, -12)
Since, p is negative, a + b has to be negative as well.
We will focus only on the negative values of the roots a and b

p = a + b

From the values of the roots above, we have p = -7, -8 and -13
The greatest integer of these three is -7
IMO, (A)
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SiffyB that’s what I did too but the OA is different.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 15 Jun 2021, 13:58
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Asitima
SiffyB
Bunuel
The roots of \(x^2 - px + 12 = 0\), where p is a negative constant and x is a variable, are a and b. If \(|a - b| \geq 12,\) what is the greatest possible integer value of p?

A. -7
B. -8
C. -13
D. -14
E. -15

Are You Up For the Challenge: 700 Level Questions

We know that,
sum of roots = a + b = p, product of roots = a * b = 12

Now, find out the values of a and b which satisfy the equation a * b = 12
They are:
(2, 6), (3, 4), (1, 12), (-2, -6), (-3, -4), (-1, -12)
Since, p is negative, a + b has to be negative as well.
We will focus only on the negative values of the roots a and b

p = a + b

From the values of the roots above, we have p = -7, -8 and -13
The greatest integer of these three is -7
IMO, (A)



SiffyB that’s what I did too but the OA is different.
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I also made the same mistake. I guess our mistake was to assume that roots are integer. With p = -14, the equation is x^2+14x+12 = 0. It has non-integer roots.
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The roots of x^2 - px + 12 = 0, where p is a negative constant and x 24 Sep 2024, 07:01
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