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D01-41

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Re: D01-41 [#permalink]

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New post 03 Jul 2016, 03:20
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?

Alternate method:

Concept: General form of Quadratic Equation = x^2 +Bx +C

Sum of roots = -B
Product of roots = C

From second equation we get product of roots = 15 and one of the roots is 3
Therefore second root must be 5
Sum = 8
Therefore a = -8

Sum of roots is common to both equations (a)
Roots are equal for first equation

2 * root = -8
Therefore root of equation 1 = -4

Product of roots is 16

Hence equation becomes x^2+ax+16=0

Therefore b = -16
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Re: D01-41 [#permalink]

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New post 10 Jul 2016, 02:53
d=(−8)2+4b=0d=(−8)2+4b=0 . Solving for bb gives b=−16b=−16.
Not clear as to why we are placing value of a in the above equation.
Since equal roots,as per what I understand ,we should be placing the value in this equation.b^2-4ac=0
But I know that there is some other approach to it and I am unclear about that.
Please help
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Re: D01-41 [#permalink]

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New post 20 Jul 2016, 19:19
Bunuel Why is the discriminant zero if it has equal roots?
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Re: D01-41 [#permalink]

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New post 21 Jul 2016, 03:01
manik919 wrote:
Bunuel Why is the discriminant zero if it has equal roots?


Check below links for some theory on algebra:
Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Theory on Algebra: algebra-101576.html
Algebra - Tips and hints: algebra-tips-and-hints-175003.html

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html

Hope this helps.
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Re: D01-41 [#permalink]

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New post 16 Apr 2017, 13:07
Bunuel wrote:

Expression \(b^2-4ac\) is called discriminant:
  • If discriminant is positive quadratics has two roots;
  • If discriminant is negative quadratics has no root;
  • If discriminant is zero quadratics has one root.

Check more HERE.



Great explanation, thank you
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Re: D01-41 [#permalink]

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New post 30 Aug 2017, 06:40
Bunuel wrote:
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. \(-\frac{1}{64}\)
B. \(-\frac{1}{16}\)
C. \(-15\)
D. \(-16\)
E. \(-64\)


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


Why you substitute the -8 back into the first equation? what is the relation between the two equations?
I know my question might be stupid but I am not getting what they are asking, maybe it´s my english...
thanks!!
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Re: D01-41 [#permalink]

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New post 16 Jan 2018, 11:11
I don't really know where I am going wrong.

\(x^2+ax−b=0\) has equal roots, and one of the roots of the equation \(x^2+ax+15=0\) is 3

\(3^2 + 3a + 15 = 0\)

\(3a = -15 - 9\)

\(3a = -24\)

\(a = -8\) Substituting it in 1st equation

\(x^2-8x-b=0\) because we know it has equal roots, we can put x = 3 in equation 1
\(9-8(3)-b=0\)
\(-b = -9+24\)
\(b = -15\)

Isn't this correct?
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Re: D01-41 [#permalink]

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New post 16 Jan 2018, 21:23
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vinod94 wrote:
I don't really know where I am going wrong.

\(x^2+ax−b=0\) has equal roots, and one of the roots of the equation \(x^2+ax+15=0\) is 3

\(3^2 + 3a + 15 = 0\)

\(3a = -15 - 9\)

\(3a = -24\)

\(a = -8\) Substituting it in 1st equation

\(x^2-8x-b=0\) because we know it has equal roots, we can put x = 3 in equation 1
\(9-8(3)-b=0\)
\(-b = -9+24\)
\(b = -15\)

Isn't this correct?


\(x^2 + ax - b = 0\) has equal roots, means that its discriminant is zero. For example, (x - 3)^2 = 0, (2x + 1)^2 = 0, ... so basically any (ax - b)^2 = 0 has has equal roots: two same roots.

\(x^2 + ax - b = 0\) has equal roots does not mean that it has equal root with \(x^2 + ax + 15 = 0\), so x = 3 might not be the root of \(x^2 + ax - b = 0\).
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D01-41 [#permalink]

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New post 06 Jun 2018, 15:30
Since the roots in the 1st equation are equal, b has to be a perfect square.

So eliminate A, B and C.

Between D and E, try to plug in b= -16 then it will be (x - 4) (x - 4) = x^2 - 8x +16

That's a match! Answer choice D

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Re: D01-41 [#permalink]

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New post 10 Jun 2018, 10:23
This is a tricky one. Since the 1st equation has equal roots, "b" has to be a perfect square.
Eliminate A, B and C.

Between D and E, plug in b=-16 then we'll have (x-4)^2 that will match x^2 - 8x + 16 = 0.

Thanks!
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Re D01-41 [#permalink]

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New post 17 Jun 2018, 07:44
I don't agree with the explanation. by subsituting a= -8 and x = 3, b= -15
Re D01-41   [#permalink] 17 Jun 2018, 07:44

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