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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] 30 Aug 2017, 06:40

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