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The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where
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10 Apr 2020, 01:37
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Competition Mode Question 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 Are You Up For the Challenge: 700 Level Questions
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where
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10 Apr 2020, 02:51
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
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10 Apr 2020, 03:29
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
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10 Apr 2020, 03:47
\(b= (\)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
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10 Apr 2020, 04:44
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) Posted from my mobile device



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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where
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10 Apr 2020, 06:34
Since roots of x^2+bx+c=0 are 3a and 4a ,then x= 3a or 4a Asked b^2 +c =? So (x3a)(x4a) =0 x^23ax4ax+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 Posted from my mobile device
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Re: The quadratic equation x^2 + bx + c = 0 has two roots 4a and 3a, where
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10 Apr 2020, 11:39
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
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10 Apr 2020, 23:03
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
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11 Apr 2020, 00:03
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
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11 Apr 2020, 01:07
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
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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
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11 Apr 2020, 02:51
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
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11 Apr 2020, 08:54
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
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12 Apr 2020, 04:10
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
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12 Apr 2020, 06:10
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
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Updated on: 13 Apr 2020, 01:22
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 Posted from my mobile device
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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
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12 Apr 2020, 10:19
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
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12 Apr 2020, 17:42
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
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12 Apr 2020, 18:22
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
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12 Apr 2020, 21:55
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
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12 Apr 2020, 21:55



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