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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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New post 10 Apr 2020, 01:37
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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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New post 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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New post 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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New post 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 \(a=3\)
\(61*9 = 549\)

The answer is therefore 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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New post 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

FINAL ANSWER IS (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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New post 10 Apr 2020, 06:34
2
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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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where  [#permalink]

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New post 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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New post 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....

Answer = 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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New post 11 Apr 2020, 00:03
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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

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.
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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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New post 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
Option C is the answer

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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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New post 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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New post 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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New post 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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New post 12 Apr 2020, 04:10
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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!
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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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New post 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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New post 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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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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New post 12 Apr 2020, 10:19
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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.
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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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New post 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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New post 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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New post 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.
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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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