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# If a1 and a2 are the real roots of x^2-px+12=0, then which of the

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Intern
Joined: 29 Jul 2019
Posts: 4
If a1 and a2 are the real roots of x^2-px+12=0, then which of the  [#permalink]

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Updated on: 05 Oct 2019, 01:09
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If $$a1$$ and $$a2$$ are the real roots of $$x^2-px+12=0$$, then which of the following statements is definitely true?

a) |$$a1+a2$$| $$\leq$$ 2$$\sqrt{3}$$
b) |$$a1-a2$$| $$\leq$$ 2$$\sqrt{3}$$
c) |$$a1+a2$$| $$\geq$$ 4$$\sqrt{3}$$
d) |$$a1-a2$$| $$\geq$$ 4$$\sqrt{3}$$
e) None of the above

Source: CAT 2019 India

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Originally posted by darshshah981 on 05 Oct 2019, 00:02.
Last edited by darshshah981 on 05 Oct 2019, 01:09, edited 3 times in total.
##### Most Helpful Community Reply
Intern
Joined: 29 Jul 2019
Posts: 4
If a1 and a2 are the real roots of x^2-px+12=0, then which of the  [#permalink]

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05 Oct 2019, 00:28
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darshshah981 wrote:
If $$a1$$ and $$a2$$ are the real roots of $$x^2-px+12=0$$, then which of the following statements is definitely true?

a) |$$a1+a2$$| $$\leq$$ 2$$\sqrt{3}$$
b) |$$a1-a2$$| $$\leq$$ 2$$\sqrt{3}$$
c) |$$a1+a2$$| $$\geq$$ 4$$\sqrt{3}$$
d) |$$a1-a2$$| $$\geq$$ 4$$\sqrt{3}$$
e) None of the above

$$x^2-px+12=0$$

Lets look at this quadratic equation is the form of $$ax^2+bx+c=0$$
We know that sum of the roots of a quadratic is $$-b/a$$ and the product of the roots of a quadratic is $$c/a$$

Now, if the roots are real then the discriminant $$\geq$$ 0, that is $$b^2 - 4ac$$ $$\geq$$ 0

>> $$p^2 - 4(1)(12)$$ $$\geq$$ 0
>> $$p^2$$ $$\geq$$ 48
>> |$$p$$| $$\geq$$ $$\sqrt{48}$$
>> |$$p$$| $$\geq$$ $$4\sqrt{3}$$

But p is the sum of the roots (a1+a2)...(Since sum of the roots of a quadratic equation is $$-b/a$$)

Therefore, |$$a1+a2$$| $$\geq$$ 4$$\sqrt{3}$$ or C is the correct answer.

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##### General Discussion
Director
Joined: 28 Jul 2016
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Re: If a1 and a2 are the real roots of x^2-px+12=0, then which of the  [#permalink]

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05 Oct 2019, 08:20
sum of roots should be greater than 0
hence $$p^2$$>48
or $$p>4\sqrt{3}$$
now p = sum of roots
hence C is the correct answer
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GRE 1: Q170 V167
If a1 and a2 are the real roots of x^2-px+12=0, then which of the  [#permalink]

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05 Oct 2019, 11:04
darshshah981 wrote:
If $$a1$$ and $$a2$$ are the real roots of $$x^2-px+12=0$$, then which of the following statements is definitely true?

a) |$$a1+a2$$| $$\leq$$ 2$$\sqrt{3}$$
b) |$$a1-a2$$| $$\leq$$ 2$$\sqrt{3}$$
c) |$$a1+a2$$| $$\geq$$ 4$$\sqrt{3}$$
d) |$$a1-a2$$| $$\geq$$ 4$$\sqrt{3}$$
e) None of the above

Source: CAT 2019 India

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In the GMAT, there is almost always a way to solve questions regarding quadratics without using the quadratic formula.

For all quadratics in this form $$x^2 - ax + b = 0$$, $$a$$ is the sum of roots and $$b$$ is the product of roots. Therefore we know $$a1*a2 = 12$$ and $$a1 + a2 = p$$.
If both roots are positive, we can minimize $$p$$ when $$a1 = a2 = \sqrt{12}$$, so that $$p = a1 + a2 = 2\sqrt{12} = 4\sqrt{3}$$, which is a minimum so we have $$a1 + a2 >= 4\sqrt{3}$$. We can do the opposite with negative roots to get a similar result, regardless all signs point to C.
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If a1 and a2 are the real roots of x^2-px+12=0, then which of the   [#permalink] 05 Oct 2019, 11:04
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# If a1 and a2 are the real roots of x^2-px+12=0, then which of the

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