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M28-08

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M28-08  [#permalink]

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New post 16 Sep 2014, 01:28
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A
B
C
D
E

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  55% (hard)

Question Stats:

63% (01:28) correct 37% (01:27) wrong based on 169 sessions

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Re M28-08  [#permalink]

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New post 16 Sep 2014, 01:28
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Official Solution:

The equation \(x^2 + ax - b = 0\) has equal roots, and one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3. What is the value of \(b\)?

A. \(-64\)
B. \(-16\)
C. \(-15\)
D. \(-\frac{1}{16}\)
E. \(-\frac{1}{64}\)


Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^3+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).


Answer: B
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Re: M28-08  [#permalink]

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New post 04 Nov 2014, 12:38
I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

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Re: M28-08  [#permalink]

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New post 05 Nov 2014, 05:28
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JackSparr0w wrote:
I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

Thanks


Yes, the discriminant of \(ax^2 + bx + c = 0\) is \(b^2 - 4ac\).

Thus the discriminant of \(x^2 - 8x - b = 0\) is : \((-8)^2 - 4(-b) = (-8)^2 + 4b\).

Hope it's clear.
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Re: M28-08  [#permalink]

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New post 05 Nov 2014, 07:58
Got it, thanks!
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Re: M28-08  [#permalink]

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New post 25 Dec 2014, 11:22
Isn't it just more simple to acknowledge that x^2-8x-b=0 and then, knowing the roots are same, use reverse foil method find: 4+4=8, 4*4=16. ??

Knowing that Bunuel always uses the simplest method, please explain the importance of knowing the discriminant? I thought with reverse foil the discriminant was not important to remember.

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M28-08  [#permalink]

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New post 21 Mar 2015, 10:50
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This question was duplicated from Q D01-41. I got this question twice from a quiz I was doing.
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Re: M28-08  [#permalink]

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New post 18 Oct 2018, 21:44
Alternative method:


We know one of the roots of x^2 +ax + 15 = 0, so let x=3
a=-8 . . . . . by substituting x=3 in x^2 +ax + 15 = 0

Standard form of quadratic equation is ax^2+bx+c=0, let its root be p and q. so, (x-p)(x-q)=0
Solving (x-p)(x-q)=0, we get x^2 -(p+q)+pq=0


By comparing x^2 -(p+q)+pq=0 with x^2 +ax - b = 0 that is x^2 -8x - b=0 . . . . .as a=-8
we can deduce that p+q=8 and pq=-b
It is given that the roots are equal so both the roots p and q must be 4 as p+q=8
and pq=16

Substituting above values in pq=-b we get that -b=16 so b=-16


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Re: M28-08  [#permalink]

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New post 18 Oct 2018, 23:23
We know one of the roots of x^2 +ax + 15 = 0 is 3.
Substitute x=3 in the equation.
We get x= -8 as the other root.
We know that sum of roots in equation ax^2 +bx + c = 0, is -b/a and product of the roots is c/a.
So, in equation x^2 +ax -b = 0, a becomes -8 => equation becomes, x^2 -8x - b = 0
(Let root 1= root 2 = r)
r + r = -8 => r = -4
r*r = -b => b = -16

Ans B

I was comfortable using the sum and product of roots concept and could solve faster that way. Hence, I thought I might add this as a solution.

Cheers!
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Re: M28-08   [#permalink] 18 Oct 2018, 23:23
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