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Re: A author writes in his book : "I was born in year (x^2 + 4x + 4 ) and [#permalink]
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Donnie84 wrote:
Hi Senthil1981,

I got up to k = (x+2)(x+3) but then picked E.

How have you concluded that only one value of x between 1800 and 1850 satisfies the equation?



Hi Donnie84, We know \(40^2 = 1600\), so i picked 42 and 43 and applied in the equation and (x+2) = 42 satisfied the condition.
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Re: A author writes in his book : "I was born in year (x^2 + 4x + 4 ) and [#permalink]
Senthil1981 wrote:
Donnie84 wrote:
Hi Senthil1981,

I got up to k = (x+2)(x+3) but then picked E.

How have you concluded that only one value of x between 1800 and 1850 satisfies the equation?



Hi Donnie84, We know \(40^2 = 1600\), so i picked 42 and 43 and applied in the equation and (x+2) = 42 satisfied the condition.


Interesting technique. Made sense.
+1 to you.
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Re: A author writes in his book : "I was born in year (x^2 + 4x + 4 ) and [#permalink]
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Donnie84 wrote:
Senthil1981 wrote:
Donnie84 wrote:
Hi Senthil1981,

I got up to k = (x+2)(x+3) but then picked E.

How have you concluded that only one value of x between 1800 and 1850 satisfies the equation?



Hi Donnie84, We know \(40^2 = 1600\), so i picked 42 and 43 and applied in the equation and (x+2) = 42 satisfied the condition.


Interesting technique. Made sense.
+1 to you.


Yes, if we got k = (x+2)(x+3) , we know that k is a product of tow consecutive integers. So, only 42*43 will be such a product that would be > 1800 but less than 1850. Hence, C.
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Re: A author writes in his book : "I was born in year (x^2 + 4x + 4 ) and [#permalink]
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Re: A author writes in his book : "I was born in year (x^2 + 4x + 4 ) and [#permalink]
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