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# if x^2+ax+b=(x+r)^2 , what is the value of r ? 1)

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if x^2+ax+b=(x+r)^2 , what is the value of r ? 1) [#permalink]  22 Feb 2005, 06:08
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if $$x^2+ax+b=(x+r)^2$$, what is the value of r ?

1) b=9

2) a=-6
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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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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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Re: DS - Equality [#permalink]  22 Feb 2005, 09:26
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)

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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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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
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"No! Try not. Do. Or do not. There is no try.

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Yup we assume x is a variable. The answer will be (B) for unique answer
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Re: DS - Equality [#permalink]  10 Sep 2009, 05:36
Sorry that I'm digging up some pretty old question, but it has totally confused me....

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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Re: DS - Equality [#permalink]  16 Sep 2009, 10:01
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.
Re: DS - Equality   [#permalink] 16 Sep 2009, 10:01
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