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Updated on: 12 May 2020, 01:37
Originally posted by LakerFan24 on 23 Apr 2017, 07:46.
Last edited by Bunuel on 12 May 2020, 01:37, edited 2 times in total.
Last edited by Bunuel on 12 May 2020, 01:37, edited 2 times in total.
Renamed the topic and edited the question.
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C
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Difficulty:
45%
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Question Stats:
71% (02:23) correct
29%
(02:25)
wrong
based on 363
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If -4 is one solution to the equation x^2 + bx + c = 0, where b and c are constants, and b + c = -21, what is the other solution?
A) -5
B) -1
C) 5
D) 7
E) 10
A) -5
B) -1
C) 5
D) 7
E) 10
23 Apr 2017, 07:59
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x^2 + bx + c =0 and -4 is one of the solution => 16+(-4)b+c=0=> 4b-c=16 and given b+c =-21; solving simultaneous equations would give b=-1
now, sum of the roots = -b/1 =>-4 + other root = 1 => other solution = 5 => option C
now, sum of the roots = -b/1 =>-4 + other root = 1 => other solution = 5 => option C
23 Apr 2017, 08:04
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Hi LakerFan24
There are a number of methods to solve this one.
But let me iterate a method that will be helpful in solving some difficult questions as well.
So for the equation x^2+bc+c => Product of roots => c/1 =c
One of the roots =-4
Hence putting x=-4 in the equation => 4b-c=16
Also we are told that b+c=21
Hence b=-1 and c=-20
Thus product of roots =-20
One of them is -4
Let the other one be G
G*(-4)=-20
G=5
Now, there are other methods which are 10000X times simpler then this one.
But as i said => This concept of sum and product of roots is helpful in some difficult questions as well.
Regards
Stone Cold
There are a number of methods to solve this one.
But let me iterate a method that will be helpful in solving some difficult questions as well.
Quote:
So for the equation x^2+bc+c => Product of roots => c/1 =c
One of the roots =-4
Hence putting x=-4 in the equation => 4b-c=16
Also we are told that b+c=21
Hence b=-1 and c=-20
Thus product of roots =-20
One of them is -4
Let the other one be G
G*(-4)=-20
G=5
Now, there are other methods which are 10000X times simpler then this one.
But as i said => This concept of sum and product of roots is helpful in some difficult questions as well.
Regards
Stone Cold
