How many integers can let x^2+bx+c=0? B and c are constants.

Question

How many integers can let x^2+bx+c=0? B and c are constants. 
1) c<0 
2) b= - (c+1)

iced_tea · Answer

B. (assuming B and C are integers) 
If B and C are not integers than E.

roots x1 = (-b + (b^2-4c)^1/2)/2 and x2 = (-b + (b^2-4c)^1/2)/2

1) c < 0 - insufficient - there could be integer as well as non integer roots.

2) gives
x1 = (-b + (c^2 + 1 + 2c -4c)^1/2)/2
 = (c+1 + (c-1))/2 = c

similarly, x2 = (c+1 -(c-1))/2 = 1

so both are integer roots (Sufficient)

ps_dahiya · Answer

E

Roots are -b/2+SQRT(b^2 - 4c)/2 and -b/2 - SQRT(b^2 - 4c)/2

St1: c < 0
then b^2 - 4c is +ve and there are two roots but we don't know if these will be integers or not.: INSUFF

St2: we have two roots
x = c and 1
so we can have either 1 root if c = 1 and two roots if c is not equal to 1.: INSUFF

Together:
Still INSUFF

haas_mba07 · Answer

Answer: B

Q: How many integer solutions for x^2 + bx + c = 0?

Let x^2 + bx +c = (x+m)(x+n) where m and n are integers (could be multiple values)

=> (m+n) = b, mn = c

S1: c < 0
If c < 0 => m < 0, n > 0 OR
If c < 0 => m > 0, n < 0

Not suffcient.

S2: b = -c -1

or m+n = -mn -1

or mn +m +(n+1) = 0
or m(n+1)+1(n+1) = 0
or (m+1)(n+1) = 0

Or m = -1, n = -1

Or b = -2, c = 1

Sufficient to find solution for
x^2 -2x+1 = (x-1)^2 = 0

Or x = 1.

One integer solution.

Sufficient.

Answer: B

iced_tea · Answer

i think you are right..i misread the question and thought it is testing whether the equation has integer roots ....

Futuristic · Answer

This is a quadratic equation. It may have 0, 1 or 2 integer roots.

x =  /2a
=  /2

From 1:

c < 0 

Now b^2 is +ve
so b^2 - 4c is +ve. This implies there are 2 roots to the equation. Are they both integers? Is one an integer? We don't know. INSUFF

From 2:

b =(-c+1)

b^2 -4c
=c^2+1+2c-4c
=(c-1)^2

Then the roots of x are
 /2 and  /2

=c, 1

Since we only know that c is a constant, but not that its an integer...INSUFF

Taking 1, 2 together...we still don't know if c is an integer....hence E

How many integers can let x^2+bx+c=0? B and c are constants.

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