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# D01-41

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Manager
Joined: 21 Sep 2015
Posts: 82

Kudos [?]: 137 [0], given: 323

Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41
Re: D01-41 [#permalink]

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03 Jul 2016, 02:20
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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Manager
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GMAT 1: 580 Q37 V33
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GMAT 5: 680 Q45 V37
GMAT 6: 690 Q47 V37
Re: D01-41 [#permalink]

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10 Jul 2016, 01:53
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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20 Jul 2016, 18:19
Bunuel Why is the discriminant zero if it has equal roots?

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Re: D01-41 [#permalink]

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21 Jul 2016, 02:01
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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16 Apr 2017, 12:07
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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Intern
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Re: D01-41 [#permalink]

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30 Aug 2017, 05:40
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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16 Jan 2018, 10:11
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] 16 Jan 2018, 10:11

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# D01-41

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