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Math Expert V
Joined: 02 Sep 2009
Posts: 58378
Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 61% (02:28) correct 39% (01:55) wrong based on 103 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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NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1022
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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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Intern  S
Joined: 22 May 2018
Posts: 49
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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Can anybody provide solution for this please?
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Manager  B
Joined: 17 Sep 2017
Posts: 101
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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Hi, can anyone provide the solution?
Senior Manager  S
Joined: 12 Sep 2017
Posts: 301
Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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
NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1022
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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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Manager  S
Joined: 15 Jan 2018
Posts: 60
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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Is there a solution that doesn't use the quadratic formula?

Bunuel
Manager  B
Joined: 27 Oct 2017
Posts: 72
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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Hi Karishma/Bunnel

Can you guys help with this?

Regards
Nidhi
Intern  B
Joined: 24 Jun 2018
Posts: 35
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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
Manager  B
Joined: 27 Oct 2017
Posts: 72
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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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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

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Director  P
Joined: 24 Nov 2016
Posts: 588
Location: United States
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

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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)

$$(x-a)(x-b)=0…x^2-xb-ax+ab=0…x^2-(a+b)x+ab=0$$
$$x^2 + [a]x + [b] = 0…[b]=ab…a=1…[a]=-(a+b)…1=-1-b…b=-2$$
$$(a,b)=(1,-2)$$ Re: Suppose that a and b are nonzero real numbers, and that the equation x   [#permalink] 13 Sep 2019, 06:42
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Suppose that a and b are nonzero real numbers, and that the equation x

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