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

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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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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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 $$b=-16$$.

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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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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16 Jan 2018, 20:23
1
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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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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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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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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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!
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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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29 Jul 2018, 01:21
1
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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29 Jul 2018, 03:15
2
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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30 Jul 2018, 03:32
1
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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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 $$b=-16$$.

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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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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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 $$b=-16$$.

ephes Feel mistake in highlighted part
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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.
Re: D01-41   [#permalink] 06 Oct 2018, 08:38

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