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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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A. -7
B. -8
C. -13
D. -14
E. -15
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Bunuel
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.
Bunuel
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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“If you are not moving forward, you are moving backward.”
General Discussion
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
\(x^2+qx+12 = 0,\) 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
Posts: 2
kris19
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?
NoobyMcScooby
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!
_________________
“If you are not moving forward, you are moving backward.”
Posts: 2
Yup, that was the mistake I was making. Thank you so much man.
Posts: 28
SiffyB
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.
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.
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.
Asitima
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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