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# Algebra question

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Intern
Joined: 15 Oct 2017
Posts: 16

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01 Jan 2018, 08:08
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Hello everyone,

i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?
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Joined: 22 May 2016
Posts: 1260

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03 Jan 2018, 20:23
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thegame12 wrote:
Hello everyone,

i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?

Attachment:

extrastepsaddedalgebra.png [ 17.59 KiB | Viewed 299 times ]

Hi thegame12 ,

A couple of steps in your screenshot were not shown.

Does it make sense now?
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At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

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Joined: 25 Dec 2017
Posts: 4
Re: Algebra question [#permalink]

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11 Jan 2018, 09:22
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thegame12 wrote:
Hello everyone,

i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?

Last one is based on (a + b)^2 = a^2 + 2ab + b^2
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11 Jan 2018, 16:40
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thegame12 wrote:
Hello everyone,

i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?

thegame12 , in my earlier reply, I misunderstood your question.

vinod94 , whose answer is very close, noticed the real question. +1 for that.

The third step to fourth step, however, comes from an algebraic identity called "the Square of a Difference," not the "Square of a Sum."*
Both are important algebraic identities with which testers should be familiar.

If an expression looks like $$(a - b)^2$$ (the square of a difference), chances are good that the expression is an important algebraic identity

In this case, $$(1 -\sqrt{b})^2$$ is an instance of the general "Square of a Difference" identity.

General identity for the square of a difference:

$$(a - b)^2 = a^2 - 2ab + b^2$$

$$(a - b)^2 = (a - b)(a - b)$$ FOIL
$$a^2 - (ab) - (ba) + b^2 =$$
$$a^2 - 2ab + b^2$$

In this case, replace $$1$$ for $$a$$, and $$\sqrt{b}$$ for $$b$$
The answer will take the form of $$a^2 - 2ab + b^2$$

$$(1 - \sqrt{b})^2 = 1 - 2\sqrt{b} + b$$

$$(1 - \sqrt{b})^2 = (1 -\sqrt{b})(1 -\sqrt{b})$$ FOIL

$$1^2 - (1)(\sqrt{b})-(1)(\sqrt{b}) + (\sqrt{b})(\sqrt{b}) =$$

$$1 - 2\sqrt{b} + b^2$$

Often a problem will have an algebraic identity "hiding."
If you recognize the identity, you can plug occasionally mind-boggling values into the identity and solve.
Three of the most important identities are listed below.

Difference of Squares:
$$(a^2 - b^2) = (a + b)(a - b)$$
Foil RHS to get LHS

Square of a Sum:
$$(a + b)^2 = a^2 + 2ab + b^2$$
$$(a + b)^2 = (a + b)(a + b)$$
FOIL RHS here to get RHS immediately above

Square of a Difference:
$$(a - b)^2 = a^2 - 2ab + b^2$$
$$(a - b)^2 = (a - b)(a - b)$$ FOIL RHS here to get RHS immediately above

Bunuel Algebraic Identities
• great article by GMAT Club expert Mike McGarry, mikemcgarry , Three Algebra Formulas Essential for the GMAT
mmagyar , two problems involving difference of squares
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At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 458
GMAT 1: 790 Q51 V49
GRE 1: 340 Q170 V170
Re: Algebra question [#permalink]

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14 Jan 2018, 13:39
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Expert's post
thegame12 wrote:
Hello everyone,

i don't get it how we can't switch from the 3rd equation to the last one. √b²=b, no ?

Interesting question. First of all, I'd say that you can switch from the third equation to the last one. You'd use FOIL (or a special quadratic that you have memorized). The earlier posters showed how that works.

If you want to double check that on your own, try plugging some numbers into both equations. If you get different results from the two equations, then you'd know that they weren't actually the same. But if you always get the same result from both of them, that would make you think that they're probably the same.

For example, try plugging in b = 0.

$$a = (1-\sqrt{0})^2 = 1^2 = 1$$

$$a = 1-2\sqrt{0}+0 = 1-2(0)+0 = 1$$

Okay, it's the same in that case. How about b = 1?

$$a = (1-\sqrt{1})^2 = 0^2 = 0$$

$$a = 1-2\sqrt{1}+1 = 1-2(1)+1 = 0$$

It's the same in that case too. What if b = 100?

$$a = (1-\sqrt{100})^2 = (-9)^2 = 81$$

$$a = 1-2\sqrt{100}+100 = 1-2(10)+100 = 81$$

It's still the same.

There's something slightly odd going on here though, which is what I think you may be asking about. Notice that I'm not plugging in any negative values for b. If I did that, I would have a problem. But the problem isn't that the two equations are different. It's just that (on the GMAT) you can't take the square root of a negative number at all. So, we don't really need to worry about negative numbers here. The GMAT will never give you expressions like these ones and ask you to use negative values with them.

Another, similar issue - but not the issue on this problem - happens when you do $$\sqrt{x^2}$$. The GMAT has a rule that says that we only look at positive square roots. So, $$\sqrt{100}$$ is just 10, not '10 or -10', like you may have learned in school. Since we don't know whether $$x$$ is positive or negative, we can't just say that $$\sqrt{x^2} = x$$. (The problem is that $$x$$ might be a negative number, which would make the equation wrong.) Instead, you have to take the absolute value of x to make sure it comes out positive: $$\sqrt{x^2} = |x|$$
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Re: Algebra question   [#permalink] 14 Jan 2018, 13:39
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# Algebra question

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