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Re: D0141
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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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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^24ac=0 But I know that there is some other approach to it and I am unclear about that. Please help



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



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16 Apr 2017, 13:07
Bunuel wrote: Expression \(b^24ac\) 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: D0141
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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: \(x^28xb=0\). 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: D0141
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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
\(x^28xb=0\) because we know it has equal roots, we can put x = 3 in equation 1 \(98(3)b=0\) \(b = 9+24\) \(b = 15\)
Isn't this correct?



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16 Jan 2018, 21:23
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
\(x^28xb=0\) because we know it has equal roots, we can put x = 3 in equation 1 \(98(3)b=0\) \(b = 9+24\) \(b = 15\)
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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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! Alecita



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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 (x4)^2 that will match x^2  8x + 16 = 0. Thanks! Alecita



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17 Jun 2018, 07:44
I don't agree with the explanation. by subsituting a= 8 and x = 3, b= 15



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11 Jul 2018, 16:31
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 (x4) (x4) = x^2  8x + 16 which match a=8 Thanks! Alecita



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23 Jul 2018, 02:08
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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Re: D0141
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29 Jul 2018, 02:21
Bunuel chetan2u niks18 gmatbusters KarishmaBQuote: 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 KarishmaBQuote: 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 \(x^28xb=0\).. it can be written in \(x^22xy+y^2=0\) to get \((xy)^2\) and thus equal roots.. So \(x^22*4*x+4^2=0\) so \(b=4^2\)
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Re: D0141
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30 Jul 2018, 04:32
adkikani wrote: Bunuel chetan2u niks18 gmatbusters KarishmaBQuote: 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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Re: D0141
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14 Aug 2018, 08:23
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: \(x^28xb=0\). 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^28x(16)=0, which is x^28x+16=0 and does not have equal roots Thanks for your help,







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