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19 Mar 2019, 00:18
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:15) correct
33%
(01:55)
wrong
based on 331
sessions
History
Date
Time
Result
Not Attempted Yet
Suppose that a and b are nonzero real numbers, and that the equation x^2 + ax + b = 0 has solutions a and b. Then the pair (a,b) is
(A) (-2, 1)
(B) (-1, 2)
(C) (1, -2)
(D) (2, -1)
(E) (4, 4)
(A) (-2, 1)
(B) (-1, 2)
(C) (1, -2)
(D) (2, -1)
(E) (4, 4)
18 Apr 2019, 01:43
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A much much quicker solution to the above problem is as follows:
We know that in a quadratic equation of the form ax^2+bx+c=0
Sum of roots=-b/a
Product of roots=c/a
In this question with quadratic equation of the form x^2+ax+b=0, the roots are also a & b
Hence, product of roots=b/1
a x b = b
Hence a = 1
Sum of roots = -a
a+b = -a
b=-2a
We know a =1
Hence b = -2
Hence option C
We know that in a quadratic equation of the form ax^2+bx+c=0
Sum of roots=-b/a
Product of roots=c/a
In this question with quadratic equation of the form x^2+ax+b=0, the roots are also a & b
Hence, product of roots=b/1
a x b = b
Hence a = 1
Sum of roots = -a
a+b = -a
b=-2a
We know a =1
Hence b = -2
Hence option C
19 Mar 2019, 10:34
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The equation \(x^2+ax+b = 0\) has two solutions. Means Discriminant \(b^2-4ac >0\)
Here b = a.
a = 1
c = b
Then we get \(b^2-4ac >0\) as \(a^2-4b > 0\)
From the given options, Only D satisfies.
a = 2
b = -1
Then \(a^2-4b > 0\) = \(4+4 > 0\)
D is the answer.
Here b = a.
a = 1
c = b
Then we get \(b^2-4ac >0\) as \(a^2-4b > 0\)
From the given options, Only D satisfies.
a = 2
b = -1
Then \(a^2-4b > 0\) = \(4+4 > 0\)
D is the answer.
