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D01-41

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Re: D01-41  [#permalink]

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New post 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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Re: D01-41  [#permalink]

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New post 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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New post 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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New post 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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New post 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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Re: D01-41  [#permalink]

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New post 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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New post 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]

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New post 16 Jan 2018, 20:23
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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, 14:30
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

Thanks!
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New post 10 Jun 2018, 09:23
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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New post 17 Jun 2018, 06:44
I don't agree with the explanation. by subsituting a= -8 and x = 3, b= -15
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New post 11 Jul 2018, 15:31
Since the 1st equation has equal roots, b has to be a perfect square. This way, eliminate answer choices A, B and C.

In the 2nd equation one of the roots is 3, so plug in x=3 then we'll get a=-8

Between answer choices D and E, choose -16 that will give (x-4) (x-4) = x^2 - 8x + 16 which match a=-8

Thanks!
Alecita :)
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New post 23 Jul 2018, 01:08
Allow me to clarify. You are confusing two different statements mentioned in the question about two different quadratic equations. Here check this complete and clear solution by Bunuel.
iPriya07 wrote:
I don't agree with the explanation. by subsituting a= -8 and x = 3, b= -15

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New post 29 Jul 2018, 01:21
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Bunuel chetan2u niks18 gmatbusters KarishmaB

Quote:
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?


Could this also be solved using

\(x^2\) - (Sum of roots)*x + (Product of roots)= 0

To validate my understanding, x is variable in question stem, where as a and b are constants.
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adkikani wrote:
Bunuel chetan2u niks18 gmatbusters KarishmaB

Quote:
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?


Could this also be solved using

\(x^2\) - (Sum of roots)*x + (Product of roots)= 0

To validate my understanding, x is variable in question stem, where as a and b are constants.


So let the roots in second case be 3and z, so \(3z=15...z=5\)
So sum of roots =3+5=8, a=-8
The first equation becomes \(x^2-8x-b=0\)..
it can be written in \(x^2-2xy+y^2=0\) to get \((x-y)^2\) and thus equal roots..
So \(x^2-2*4*x+4^2=0\) so \(b=-4^2\)
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Re: D01-41  [#permalink]

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New post 30 Jul 2018, 03:32
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adkikani wrote:
Bunuel chetan2u niks18 gmatbusters KarishmaB

Quote:
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?


Could this also be solved using

\(x^2\) - (Sum of roots)*x + (Product of roots)= 0

To validate my understanding, x is variable in question stem, where as a and b are constants.


Yes, as shown by chetan2u above and in fact you can continue to use the same approach in the last steps also:

\(x^2 + ax + 15 = 0\)
If one root is 3, the other is 5 (to get the product 15). So sum of the roots is 8.
This means a = -8

The other equation becomes \(x^2 - 8x - b = 0\)
The sum of the roots is 8 and the roots are equal to they will be 4 and 4. Their product becomes 16. So b becomes -16.
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New post 14 Aug 2018, 07:23
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



Dear Bunuel

Thank you for your explanations, they are always really helpful

The way the question is typed, shouldn't b be equal to +16 (and not -16)? Because the equation has already a - sign before b, if we plug in the answer -16, the equation becomes x^2-8x-(-16)=0, which is x^2-8x+16=0 and does not have equal roots

Thanks for your help,
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New post 15 Aug 2018, 18:20
I don't think it's fair to say knowing the discriminant rules are "the fundamentals".

Most test takers will have to guess and check after figuring out that a= -8
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New post 25 Sep 2018, 10:07
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


ephes Feel mistake in highlighted part
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New post 06 Oct 2018, 08:38
I solved it to the point, a=-8.

Then, used the approach sum of roots = -b/a (if the equation is of the form ax2 +bx+c)

So the sum is 8/1= 8 =2x (since the roots are same). This is how I got the value of x=4

Then after that using the method of substitution I could get the value of b.

x(square)+ ax-b=0
(4*4)+4(-8)-b=0
16-32-b=0
-16-b=0
Hence b=-16

Please let me know if my approach is correct.
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