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Bunuel
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a and b are roots of the equation x^2 - px + 12 = 0. If the difference 08 Dec 2021, 09:15
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C
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52% (02:36) correct 48% (02:26) wrong based on 82 sessions

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a and b are roots of the equation \(x^2 - px + 12 = 0\). If the difference between the roots is at least 12, what is the range of values p can take?

A. (0, 2)
B. (-2, 2)
C. (-∞, -2) or (-2, 2)
D. (-2, 0)
E. (2, 12)


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C
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DHRJ0032
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a and b are roots of the equation x^2 - px + 12 = 0. If the difference 03 Jan 2022, 11:53
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from eqn,
a+b=-p
ab=12

&
a-b>=12
using formula,
(a+b)2 -(a-b)2=4ab
we get ,
p(sqr) >=192
range of p comes out to be (-infinite,-8 sqroot 3) &(8 sqroot 3, infinite)
none of the options is matching with the above mentioned range.
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a and b are roots of the equation x^2 - px + 12 = 0. If the difference 17 Mar 2022, 05:45
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A simple way of doing this is -

x^2 - px + 12 = 0 can be solved by factorization if p = 13 or -13. Because only C contains this option, that must be the correct answer.
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a and b are roots of the equation x^2 - px + 12 = 0. If the difference 17 Mar 2022, 09:13
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Let Root 1 and Root 2 be A and B

Template:

(sum of root)^2 — (difference of roots)^2 = 4 * (product of roots)

Or

(A + B)^2 - (A - B)^2 = 4AB

If P > 0 ——-> SUM of Roots = (-) (-P) = P

Product of Roots. = A*B = 12

Substitute these 2 expressions into the template above

(P)^2 - (A - B)^2 = 4(12) = 48

(P)^2 - 48 = (A - B)^2 ——- (equation 1)


Using the given information: the difference of the Roots is at least 12

(A - B) >/= 12

—both sides of inequality are positive —- Square both sides—-

(A - B)^2 >/= 144 —— (equation 2)

Substitute (equation 1) in for the value of (A - B)^2


P^2 - 48 >/= 144

P^2 >/= 192

[P] >/= sqrt(192)

If we assume instead that P < 0 at the outset:

P </= (-1) * sqrt(192)

Or

P </= (-)14

*C* seems to be the only answer choice that contains values within this range.

Bunuel

It is this last part during which I am having difficulty conveying the algebra into a proper range of answers.

Do you have any suggestions?
Thanks.

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a and b are roots of the equation x^2 - px + 12 = 0. If the difference 16 Oct 2024, 13:57
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Bunuel, can you verify if my approach is fine ?

Let roots be a,b
a*b = 12
a+b = p
|a-b|>= 12

(a-b)^2 = a^2 + b^2 - 2ab >= 144
a^2 + b^2 >= 168

(a+b)^2 = p^2
a^2 + b^2 + 2ab = p^2
=> p^2 >= 192

Implies p >= 13.xx or p =< -13.xx

Only c contains part of the range.
Is there any other approach to solve this ?

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Bunuel
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