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Re: equations

by Morgoth » Thu Oct 02, 2008 3:28 pm
beater wrote:If x^2 + 3x + c = (x + a)(x + b) for all x, what is the value of c ?
(1) a = 1
(2) b = 2
x^2 + 3x + c = (x + a)(x + b)

here (x+a) & (x+b) are the two roots of the equation

statement (1)
a=1
x+1 = 0
x=-1
1-3+c = 0
c=2
Sufficient.

Statement (2)
b=2
x+2 = 0
x=-2
4-6+c = 0
c=2
Sufficient.

Thus, D.

OA?
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by beater » Thu Oct 02, 2008 9:00 pm
Hi Morgoth - How did you come up with this:
"here (x+a) & (x+b) are the two roots of the equation"
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by stop@800 » Thu Oct 02, 2008 9:10 pm
The roots can only eb assumed if we have
x^2 + 3x + c = (x + a)(x + b) = 0

My suggestion would be
you expand equation (x + a)(x + b)
and compare corresponding powers of x

This may turn out to be same as equating both to 0
but I think we should not assume that
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by raju232007 » Fri Oct 03, 2008 10:03 am
As stop said we can compare the powers and find the solution...

statement 1
a=1

x^2+3x+c=(x+1)(x+b)
x^2+3x+c=x^2+x(b+1)+b
comparing the powers we have
b+1=3
b=2
b=c
c=2

statement 2
b=2
x^2+3x+c=(x+a)(x+2)
x^2+3x+c=x^2+(a+2)x+2a
comparing the powers we have
a+2=3
a=1
c=2a
c=2

Hence the ans is D
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