The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where

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The quadratic equation  x^2 + bx + c = 0  has two roots  4a  and  3a , where  a  is an integer. Which of the following is a possible value of  b^2 + c  ?

A. 3721
B. 550
C. 549
D. 427
E. 361

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ArunSharma12 · Accepted Answer

The quadratic equation x^2+bx+c=0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^2+c ?

A. 3721
B. 550
C. 549
D. 427
E. 361

sum of the roots = 7a = -b
product of the roots =  12a^2  =c
 b^2=49a^2 
 b^2+c=49a^2+12a^2=61a^2

lets take a look at the options
A. 3721 - 61*61
B. 550 - not a multiple of 61
C. 549 - 61*9
D. 427 - 61*7
E. 361 - 61*6

out of the given options only option C matches  61*9 = 61*a^2 
Ans: C

GMATWhizTeam · Answer

Solution:  
 • Sum of the roots = -  \frac{Coefficient \ of \ x }{ Coefficient \ of \ x^2} 
 o  3a + 4a = -b 
o  b = -7a 
o  b^2 = 49a^2  
• Product of roots =  \frac{constant}{Coefficient \ of \ x^2} 
 o  3a*4a = \frac{c}{1} 
o  c = 12a^2  
•  b^2 + c = 49a^2 + 12a^2 = a^2*(61) 
 o  b^2 +c  is a multiplication of 61 and a perfect square, lets see which of the given option satisfy this condition.  
 Option A:  3721
If  a^2*61 = 3721 , then  a^2 = \frac{3721}{61} 
 •  a^2 = 61 , in this case a will not be an integer. Not possible 
 Option B:  550
If  a^2*61 = 550 , then  a^2 = \frac{550}{61} 
 •  a^2 = \frac{550}{61} , in this case a will not be an integer. Not possible 
 Option C:  549
If  a^2*61 = 546 , then  a^2 = \frac{549}{61} 
 •  a^2 = 9 , 
 o  a=3 , Option C must be the correct answer  
Hence, the correct answer is  Option C.

eakabuah · Answer

b= -( sum of roots )  hence  b=-7a 
 c= product of roots  = 12a^2 
 b^2+c= 49a^2+12a^2 = 61a^2 
Basically the unit digit of  61a^2  is the unit digit of  a^2 .
Looking at the answer choices, we know that  a  cannot be  2  since none of the answer choices has a unit digit of  4 .
Since option C has a unit digit of  9 ,  a=3  and  a=7  are possible. Testing with  a=3 
 61*9 = 549

The answer is therefore C.

freedom128 · Answer

-b = (4a+3a) = 7a --> b^2 = 49a^2
c=4a*3a=12a^2

b^2 + c = 61a^2

Check possible value of (b^2+c) 
A. 3721 = 61*61
B. 550 = 61*9 +1
 C. 549 = 61*9 = 61*3^2 
D. 427 = 61*7
E. 361 = 61*6

FINAL ANSWER IS (C)
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Staphyk · Answer

Since roots of x^2+bx+c=0 are 3a and 4a ,then x= 3a or 4a
Asked b^2 +c =?

So (x-3a)(x-4a) =0
x^2-3ax-4ax+12a^2=0
x^2-(7a)x+12a^2=0
Compared to x^2+bx+c=0
b=-7a ,c =12a^2
.: b^2+c = (-7a)^2+12a^2 =61a^2
Now we want an answer in the form (61•a•a)
Start with (c) 549 = 61•3•3
Bingo! 
C like Covid

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monikakumar · Answer

The quadratic equation x2+bx+c=0x2+bx+c=0 has two roots 4a4a and 3a3a, where aa is an integer. Which of the following is a possible value of b2+cb2+c ?

A. 3721
B. 550
C. 549
D. 427
E. 361

b=-(sum of roots)=> 7a,
b^2=49a^2
c=product of roots => 12a^2
b^2+c=49a^2+12a^2 => 61a^2
a^2 is square
Ans C (549/6=9)

shankgmat · Answer

sum of roots = -b
4a+3a=-b
7a=-b
b^2=49a^2

Also product of roots is c
c=12a^2

b^2+c=49a^2+12a^2=61a^2

So a is an integer, insert values of a = 1,2,3... in to equation i.e. 61*1=61,61*4=284,61*9=549,61*16=976....

Answer = C

unraveled · Answer

The quadratic equation  x^2+bx+c=0  has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of  b^2+c  ?

A. 3721
B. 550
C. 549
D. 427
E. 361

In quadratic equation  x^2+bx+c=0 , 
b = sum of roots and c = product of roots
b = 7a and c =  12a^2 
 b^2 = 49a^2  So,
 b^2 + c = 49a^2 + 12a^2 = 61a^2

If a = 3,  61a^2  = 61*9 = 549 
 Note: None of the square of an integer has unit digit 7 so D is straight out. From the rest of the option easiest to check is  3^2  and 60*60 = 3600 so A is out.

Answer C.

gurmukh · Answer

b=7a
c=12a^2
b^2 + c
49a^2 + 12a^2
=61a^2
61a^2 =549
a^2 =9
a =3
Option C is the answer

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PB12 · Answer

Sum of roots = -b:
4a + 3a = -b
b^2 = 49a^2

similarly product of roots = c
or 12a^2 = c

thus b^2 + c = 49a^2 + 12a^2 = 61a^2

as a is integers so b^2 + c can be 61, 61*4 , 61*9 ….so on
so option b ) 549 that is 61*9 is correct.

madgmat2019 · Answer

x2+bx+c=0 has two roots 4a and 3a, where a is an integer.

sum of roots = -b = 7a
product of roots =12a^2
 b2+c=49a^2+12a^2=61a^2

A. 3721.......61*61.....a^2=61.....a is not an int.....incorrect
B. 550......not a multiple of 61....incorrect
C. 549.........61*9,.......a=3.....correct
D. 427.......61*7.......a is not an int.....incorrect
E. 361......a is not an int.....incorrect

OA:C

Badari · Answer

Sum of the roots = b =(4a+3a)= 7a
Product of the roots = 12 a^2 = c

b^2 + c = 49 a^2 + 12 a^2 = 61 a^2

Given a is integer try a=1 Value = 61
Try a=2 Value = 61 * 4 =244
Try a=3 Value =61*9 549

Option C

Vidhi96 · Answer

Answer is "C"
We know in quad eq = sum of roots is -b/a = - (4a+3a)/ 1= -7a similarly product of roots c = 12a^2

This when inserted to b^2+c = 49a^2+ 12a^2 taking a^2 in common = a^2(61) only C serves the part where 9 is a square figure!

MrSengupta · Answer

Sum of two roots => 7a=-b and Product of two roots : 12a^2=c

Therefore b^2+c= 61a^2 . 61*9=549 . Therefore (C)

Kinshook · Answer

Sum of roots = 3a + 4a = 7a = -b
Product of roots = 49a^2 = b^2

c =12a^2

b^2 + c = 49a^2 + 12a^2 = 61a^2

c. 549 = 61*9 = 61 *3^2

IMO C

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minustark · Answer

Sum of the roots, 4a +3a = -b or, 7a =-b
Product of the roots ,  12a^2  = c
Now,  b^2 +c = (-7a)^2 +12a^2= 49a^2 +12a^2 = 61a^2  
The value can be 61, (61*4 = 244),(61*9= 549).
C is the answer.

NutHead · Answer

Sum of roots= -b = 7a
Product of roots = c = 12 a^2
b^2 +c = 61a^2
i.e 61a^2 should be a factor of one of the option in order to satisfy. Therefore 549 !

Drb89 · Answer

sum of roots = -b/a = -b = 4a+3a = 7a
product of roots = c/a = c = 4a*3a = 12a^2

b^2+c = (-7a)^2+12a^2 = 61a^2.

Only number which is a multiple of 61 is 549, which is 9*61.

Vaibhavvc1503 · Answer

-b/a= sum of roots= 7a (here in EQ. a=1)
-b=7a
b^2=49a^2 -------------(1)

c=4a*3a
c=12a^2------------------(2)

b^2+c = 61a^2 ----------(3)

now according to options 
only A, C,D are divisile by 61 and the factors are 61, 9, 7 respectively

as a^2= perfect sqaure
therefore a^2=9 ( as 9 is the only perfect square here)

hence option C is the right answer.

The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where

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GMAT Problem Solving (PS) Questions

The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where

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