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# 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  [#permalink]

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10 Apr 2020, 01:37
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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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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 02:51
2
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
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 03:29
2

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.

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 03:47
2
$$b= -($$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 $$a=3$$
$$61*9 = 549$$

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 04:44
3
-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

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 06:34
2
Since roots of x^2+bx+c=0 are 3a and 4a ,then x= 3a or 4a

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)
Bingo!
C like Covid

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 11:39
2
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)
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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10 Apr 2020, 23:03
2
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....

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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11 Apr 2020, 00:03
3
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.

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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11 Apr 2020, 01:07
2
b=7a
c=12a^2
b^2 + c
49a^2 + 12a^2
=61a^2
61a^2 =549
a^2 =9
a =3

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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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11 Apr 2020, 02:02
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.
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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11 Apr 2020, 02:51
2
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
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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11 Apr 2020, 08:54
2
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
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 04:10
1
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!
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 06:10
2
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)
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The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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Updated on: 13 Apr 2020, 01:22
2
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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Originally posted by Kinshook on 12 Apr 2020, 10:07.
Last edited by Kinshook on 13 Apr 2020, 01:22, edited 1 time in total.
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 10:19
2
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).
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 17:42
2
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 !
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 18:22
2
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.
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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12 Apr 2020, 21:55
2
-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.
Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where   [#permalink] 12 Apr 2020, 21:55

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