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SVP
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if x^2+ax+b=(x+r)^2 , what is the value of r ? 1) [#permalink]
 22 Feb 2005, 07:08
Question Stats:
100% (02:32) correct
0% (00:00) wrong based on 1 sessions
if x^2+ax+b=(x+r)^2, what is the value of r ?
1) b=9
2) a=-6
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Intern
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I agree with mirhaque on answer choice (E). How did you get 3 for r, thearch? The equation can be expanded to:
x^2+ax+b = x^2+2rx+r^2
I canceled out x^2 on both sides, leaving:
ax+b = 2rx+r^2
Statement (I) says b=9, but that still leaves you with 3 unknowns.
Statement (II) says a=-6, and that still leaves you with 3 unknowns.
Combinining (I) and (II) still leaves 2 unknowns.
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Director
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hornsfan2005 wrote: Combinining (I) and (II) still leaves 2 unknowns.
my reasoning was:
distinguish between two sides of the equation
so you resolve the first
x^2-6x+9=0 --> only one root because 36-4*9=0
so the solution of it is (6+-0)/2=3 (X found)
then you have
(x+r)^2=0 -> r=-3
but I'm now considering that the stem doesn't tell us that x^2-6x+9=(x+r)^2 =0 and without that you remain with two unknowns so it should be (E)
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Intern
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christoph wrote: if x^2+ax+b=(x+r)^2, what is the value of r ?
1) b=9
2) a=-6
My take: answer should be (B)
solving the equation
ax + b = 2rx + r^2
==> a = 2r and b = r^2 (x coefficient and the constant term should match on both side, else the equality sign does not hold)
Taking (1)
b = 9 = r^2
r can be +3 or -3. Hence no unique value
Taking (2)
a = -6 = 2r ==> r = -3 ==> r^2 = 9 = b
All values unique.
Hence answer obtained by taking only (2)
Ketan
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Current Student
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B it is...
from the equation
x^2+ax+b=x^2+2xr+r^2
r^2=b=9, we dont know if r = +- 3 (satement 1) insuff
ax=2xr, you can cancel the x, a=2r! (statment 2) suff!
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Senior Manager
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B is sufficient...
x^2+ax+b = (x+r)^2
If b = 9
then r = +3 or -3
so A and D are out
If a = -6 the the only choice for r is -3 therefore my answer is B
_________________
"No! Try not. Do. Or do not. There is no try.
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Manager
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Yup we assume x is a variable. The answer will be (B) for unique answer
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Tuck Thread Master
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Sorry that I'm digging up some pretty old question, but it has totally confused me....  Is the answer B correct? So we have: ax + b = 2rx + r^2 from which we deduce that: (ax = 2rx) AND (b = r^2) the last part of solution is crystal clear, but how can this deduction be correct? is it true that, if we have general equation a1x + b1 = a2x + b2 [where x is variable and a1,a2,b1,b2 some constants], then (a1x = a2x) AND (b1 = b2)? I don't get it 
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if x^2+ax+b=(x+r)^2, what is the value of r ?
1) b=9
2) a=-6
Soln: Lets first open up the given equation,
=> x^2+ax+b=(x+r)^2
=> x^2+ax+b=x^2+2rx+r^2
now since coefficients of x^2 are equal, so matching the other coefficients on both sides,
ax = 2rx => a = 2r => eq(1)
b = r^2 => eq(2)
Considering statement 1 alone,
since b = 9 and we have b = r^2
therefore r = +3 or -3
Since we have two values of r, Statement 1 alone is insufficient
Considering statement 2 alone,
since a = -6 and we have a = 2r
therefore r = -3
Since we have just one unique value of r, Statement 2 alone is sufficient
Hence, B.
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