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Suppose that a and b are nonzero real numbers, and that the equation x
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19 Mar 2019, 00:18
Question Stats:
59% (02:21) correct 41% (01:48) wrong based on 59 sessions
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Suppose that a and b are nonzero real numbers, and that the equation x^2 + ax + b = 0 has solutions a and b. Then the pair (a,b) is (A) (2, 1) (B) (1, 2) (C) (1, 2) (D) (2, 1) (E) (4, 4)
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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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19 Mar 2019, 10:34
The equation \(x^2+ax+b = 0\) has two solutions. Means Discriminant \(b^24ac >0\)Here b = a. a = 1 c = b Then we get \(b^24ac >0\) as \(a^24b > 0\) From the given options, Only D satisfies. a = 2 b = 1 Then \(a^24b > 0\) = \(4+4 > 0\) D is the answer.
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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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01 Apr 2019, 03:39
Can anybody provide solution for this please?
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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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06 Apr 2019, 23:54
Hi, can anyone provide the solution?



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Suppose that a and b are nonzero real numbers, and that the equation x
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09 Apr 2019, 19:06
Hello Chethan92! I did it the same way as you but I have a question. When you try C and D, both are greater than 0. How did you choose D? \(a^2  4b > 0\) C \((1 ,2)\)
\(1^2  4(2) > 0\)
\(9 > 0\) TrueD \((2 ,1)\)
\(2^2  4(1) > 0\)
\(8 > 0\) True



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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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09 Apr 2019, 19:14
jfranciscocuencag wrote: Hello Chethan92! I did it the same way as you but I have a question. When you try C and D, both are greater than 0. How did you choose D? \(a^2  4b > 0\) C \((1 ,2)\)
\(1^2  4(2) > 0\)
\(9 > 0\) TrueD \((2 ,1)\)
\(2^2  4(1) > 0\)
\(8 > 0\) TrueHey jfranciscocuencag, How are you? Yes, even I noticed it. But I missed it while I was answering this question. Maybe D option should be some other value as the answer is C.
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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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16 Apr 2019, 15:10
Is there a solution that doesn't use the quadratic formula? Bunuel



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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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17 Apr 2019, 07:54
Hi Karishma/Bunnel
Can you guys help with this?
Regards Nidhi



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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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18 Apr 2019, 01:43
A much much quicker solution to the above problem is as follows:
We know that in a quadratic equation of the form ax^2+bx+c=0 Sum of roots=b/a Product of roots=c/a
In this question with quadratic equation of the form x^2+ax+b=0, the roots are also a & b
Hence, product of roots=b/1 a x b = b Hence a = 1
Sum of roots = a a+b = a b=2a
We know a =1 Hence b = 2
Hence option C



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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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18 Apr 2019, 07:38
sahibr wrote: A much much quicker solution to the above problem is as follows:
We know that in a quadratic equation of the form ax^2+bx+c=0 Sum of roots=b/a Product of roots=c/a
In this question with quadratic equation of the form x^2+ax+b=0, the roots are also a & b
Hence, product of roots=b/1 a x b = b Hence a = 1
Sum of roots = a a+b = a b=2a
We know a =1 Hence b = 2
Hence option C Thanks..got it.. Posted from my mobile device



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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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19 Apr 2019, 10:02
Bunuel wrote: Suppose that a and b are nonzero real numbers, and that the equation x^2 + ax + b = 0 has solutions a and b. Then the pair (a,b) is
(A) (2, 1) (B) (1, 2) (C) (1, 2) (D) (2, 1) (E) (4, 4) We can create the equation: Since we are given that the roots are a and b, we can write: (x  a)(x  b) = 0 x^2  ax  bx + ab = 0 x^2  (a + b)x + ab = 0 Now, we compare the equation obtained above to the original equation x^2 + ax + b = 0, obtaining: a = (a + b) and b = ab. Since neither a nor b is 0, then a must be 1 so that b = ab, and we have: 1 = (1 + b) 1 = 1  b b = 2 Answer: C
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Re: Suppose that a and b are nonzero real numbers, and that the equation x
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19 Apr 2019, 10:02






