GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 29 May 2020, 00:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 64222
Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

18 Mar 2019, 23:18
00:00

Difficulty:

55% (hard)

Question Stats:

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

### HideShow timer Statistics

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)

_________________
NUS School Moderator
Joined: 18 Jul 2018
Posts: 1120
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE: Engineering (Energy and Utilities)
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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 $$a^2-4b > 0$$ = $$4+4 > 0$$

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

### Show Tags

01 Apr 2019, 02:39
Can anybody provide solution for this please?
Manager
Joined: 17 Sep 2017
Posts: 98
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

06 Apr 2019, 22:54
Hi, can anyone provide the solution?
Senior Manager
Joined: 12 Sep 2017
Posts: 306
Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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
NUS School Moderator
Joined: 18 Jul 2018
Posts: 1120
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE: Engineering (Energy and Utilities)
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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.
Manager
Joined: 15 Jan 2018
Posts: 66
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

16 Apr 2019, 14:10
Is there a solution that doesn't use the quadratic formula?

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

### Show Tags

17 Apr 2019, 06:54
Hi Karishma/Bunnel

Can you guys help with this?

Regards
Nidhi
Intern
Joined: 24 Jun 2018
Posts: 36
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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

### Show Tags

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..

Posted from my mobile device
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10590
Location: United States (CA)
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
202 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

SVP
Joined: 24 Nov 2016
Posts: 1563
Location: United States
Re: Suppose that a and b are nonzero real numbers, and that the equation x  [#permalink]

### Show Tags

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)

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

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