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B
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64% (02:19) correct
36%
(02:33)
wrong
based on 132
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If \(c\) is a constant, and \(a\) and \(b\) are solutions for \(x\) to the equation \(x^2 + 2cx + c^2 = 1\), then which of the following could be equal to \(a − b\)?
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
17 Jul 2021, 07:29
Kudos
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sjuniv32
If \(c\) is a constant, and \(a\) and \(b\) are solutions for \(x\) to the equation \(x^2 + 2cx + c^2 = 1\), then which of the following could be equal to \(a − b\)?
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
\(x^2 + 2cx + c^2 = 1………..(x+c)^2=1=1^2=(-1)^2\)
It is possible only when x+c=1 or x+c=-1
So two roots of the equation are
(i) x=1-c and
(ii) x=-1-c
One of the two is a and other b.
Thus,
a-b can be (1-c)-(-1-c)=2
Or
a-b can be -1-c-(1-c)=-2
Only II possible
B
19 Jul 2021, 10:30
Kudos
Bookmarks
sjuniv32
If \(c\) is a constant, and \(a\) and \(b\) are solutions for \(x\) to the equation \(x^2 + 2cx + c^2 = 1\), then which of the following could be equal to \(a − b\)?
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
I) 1
II) 2
III) 4
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
Rewrite the equation in standard form, \(x^2 + 2cx + c^2 - 1 = 0\).
Note from the constant part, we have \(c^2 - 1 = (c - 1)(c + 1)\). Then also note if we add up \(c - 1\) and \(c + 1\) we get exactly \(2c\), the coefficeint of the x term. So we can actually factor this equation into \((x - c + 1)(x - c - 1) = 0\). Thus the solutions in terms of c would be \(x = c - 1\) and \(x = c + 1\).
Finally, if we subtract the two solutions from each other we can only get 2 or -2.
Ans: B
