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Re: If s and t are constants and x2 + sx + 8 factors into (x + 1)(x + t), [#permalink]
Expert Reply
kkrpawkal123 wrote:
Hi rx10,
The solution is valid only when we consider the above equation as quadratic. How did u know that it's a quadratic equation in the first place?


Knowing the difference what is an equation or what is an expression is one of the basic skills for the GRE.

It is like the periodic table in your mind

Please see our chapter about equations https://gre.myprepclub.com/forum/gre-math- ... 25882.html

regards
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Re: If s and t are constants and x2 + sx + 8 factors into (x + 1)(x + t), [#permalink]
Carcass wrote:
kkrpawkal123 wrote:
Hi rx10,
The solution is valid only when we consider the above equation as quadratic. How did u know that it's a quadratic equation in the first place?


Knowing the difference what is an equation or what is an expression is one of the basic skills for the GRE.

It is like the periodic table in your mind

Please see our chapter about equations https://gre.myprepclub.com/forum/gre-math- ... 25882.html

regards


Hi Carcass,

Thanks for sharing the post, I still have the same doubt. As in the link that you shared, I referred to the image below,

Image

Point 1: The above image says that an equation HAS AN EQUAL TO ('=') SIGN IN IT. In the question x^2+sx+8 DO NOT HAVE ANY = SIGN.
Point 2: In turn, this x^2+sx+8 resembles an expression.(If we refer to the third example of Expression in the given image.

* Coming back to my doubt, when there's no equal to symbol in the question, how do we assume that x^2+sx+8 is an equation. Do we always have to assume them in GRE, whenever a second degree expression is given.
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Re: If s and t are constants and x2 + sx + 8 factors into (x + 1)(x + t), [#permalink]
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Usually yes

Moreover, we do know that the equation above has \((x + 1)(x + t)\) that are factors.

So it is implied that is =0, therefore, is an equation because even though we do have \((x + 1)(x + t)\) we know that is solved already with unknown

tell me if now is clear
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Re: If s and t are constants and x2 + sx + 8 factors into (x + 1)(x + t), [#permalink]
1
Now it's clear, thanks for the help Carcass. :)
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Re: If s and t are constants and x2 + sx + 8 factors into (x + 1)(x + t), [#permalink]
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