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Suppose that a and b are nonzero real numbers, and that the equation x

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Math Expert
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Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 18 Mar 2019, 23:18
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (02:27) correct 38% (01:55) wrong based on 111 sessions

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Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 19 Mar 2019, 09:34
2
1
2
The equation \(x^2+ax+b = 0\) has two solutions. Means Discriminant \(b^2-4ac >0\)
Here b = a.
a = 1
c = b

Then we get \(b^2-4ac >0\) as \(a^2-4b > 0\)
From the given options, Only D satisfies.
a = 2
b = -1
Then \(a^2-4b > 0\) = \(4+4 > 0\)

D is the answer.
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Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 01 Apr 2019, 02:39
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  [#permalink]

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New post 06 Apr 2019, 22:54
Hi, can anyone provide the solution?
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Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 09 Apr 2019, 18:06
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 \((1 ,-2)\)

\(1^2 - 4(-2) > 0\)

\(9 > 0\) True


D \((2 ,-1)\)

\(2^2 - 4(-1) > 0\)

\(8 > 0\) True
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Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 09 Apr 2019, 18:14
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 \((1 ,-2)\)

\(1^2 - 4(-2) > 0\)

\(9 > 0\) True


D \((2 ,-1)\)

\(2^2 - 4(-1) > 0\)

\(8 > 0\) True


Hey 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  [#permalink]

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New post 16 Apr 2019, 14:10
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  [#permalink]

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New post 17 Apr 2019, 06:54
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  [#permalink]

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New post 18 Apr 2019, 00:43
3
1
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  [#permalink]

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New post 18 Apr 2019, 06:38
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..

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Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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New post 19 Apr 2019, 09:02
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  [#permalink]

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New post 13 Sep 2019, 05:42
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)


\((x-a)(x-b)=0…x^2-xb-ax+ab=0…x^2-(a+b)x+ab=0\)
\(x^2 + [a]x + [b] = 0…[b]=ab…a=1…[a]=-(a+b)…1=-1-b…b=-2\)
\((a,b)=(1,-2)\)

Answer (C)
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Re: Suppose that a and b are nonzero real numbers, and that the equation x   [#permalink] 13 Sep 2019, 05:42

Suppose that a and b are nonzero real numbers, and that the equation x

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