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Intern  B
Joined: 14 Sep 2016
Posts: 13
Solve square root of (d)^2+6d+9?  [#permalink]

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Answer to the question should be:
Question: Square root of (d)^2+6d+9
Answer: Square root of (d+3)^2
d+3.

BUT in the book, they have given two answers:
d+3 and -(d+3)

Arent they wrong? if something is under the square root sign we always consider positive answer only, don't we? Please clarify this doubt.

If someone has Manhattan Prep 6th Edition Algebra book, this is on page 182 and the explanation is page 71.
Intern  B
Joined: 25 Sep 2018
Posts: 7
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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If you take the square roof of a number there are always 2 possible solutions:
E.g.:
(-x)^2 = x^2
x^2 = x^2
Hope this helps

Posted from my mobile device
Intern  B
Joined: 14 Sep 2016
Posts: 13
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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Reactzz wrote:
If you take the square roof of a number there are always 2 possible solutions:
E.g.:
(-x)^2 = x^2
x^2 = x^2
Hope this helps

Posted from my mobile device

I know there are 2 possible solutions but when the square root sign is there, there is only one positive solution as mentioned in the book. Please see my doubt.
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Joined: 31 Jul 2017
Posts: 512
Location: Malaysia
Schools: INSEAD Jan '19
GMAT 1: 700 Q50 V33 GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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aj3001 wrote:
Answer to the question should be:
Question: Square root of (d)^2+6d+9
Answer: Square root of (d+3)^2
d+3.

BUT in the book, they have given two answers:
d+3 and -(d+3)

Arent they wrong? if something is under the square root sign we always consider positive answer only, don't we? Please clarify this doubt.

If someone has Manhattan Prep 6th Edition Algebra book, this is on page 182 and the explanation is page 71.

As I understand, anything under the Square Root sign is considered as a positive.
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Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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aj3001 wrote:
Answer to the question should be:
Question: Square root of (d)^2+6d+9
Answer: Square root of (d+3)^2
d+3.

BUT in the book, they have given two answers:
d+3 and -(d+3)

Arent they wrong? if something is under the square root sign we always consider positive answer only, don't we? Please clarify this doubt.

If someone has Manhattan Prep 6th Edition Algebra book, this is on page 182 and the explanation is page 71.

$$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. So, the answer is d + 3 or -(d + 3). Notice that -(d + 3) is not necessarily negative, for example consider d = -4.
_________________
Intern  B
Joined: 14 Sep 2016
Posts: 13
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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Bunuel wrote:
aj3001 wrote:
Answer to the question should be:
Question: Square root of (d)^2+6d+9
Answer: Square root of (d+3)^2
d+3.

BUT in the book, they have given two answers:
d+3 and -(d+3)

Arent they wrong? if something is under the square root sign we always consider positive answer only, don't we? Please clarify this doubt.

If someone has Manhattan Prep 6th Edition Algebra book, this is on page 182 and the explanation is page 71.

$$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. So, the answer is d + 3 or -(d + 3). Notice that -(d + 3) is not necessarily negative, for example consider d = -4.

Thank you so much for your explanation. I got confused since in the book they have mentioned that if a square root sign is used in the GMAT, only the positive value is considered. e.g.. $$\sqrt{16}$$ then answer should be only 4 and not both 4 and -4.
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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aj3001 wrote:
Bunuel wrote:
aj3001 wrote:
Answer to the question should be:
Question: Square root of (d)^2+6d+9
Answer: Square root of (d+3)^2
d+3.

BUT in the book, they have given two answers:
d+3 and -(d+3)

Arent they wrong? if something is under the square root sign we always consider positive answer only, don't we? Please clarify this doubt.

If someone has Manhattan Prep 6th Edition Algebra book, this is on page 182 and the explanation is page 71.

$$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. So, the answer is d + 3 or -(d + 3). Notice that -(d + 3) is not necessarily negative, for example consider d = -4.

Thank you so much for your explanation. I got confused since in the book they have mentioned that if a square root sign is used in the GMAT, only the positive value is considered. e.g.. $$\sqrt{16}$$ then answer should be only 4 and not both 4 and -4.

The book is correct.

Check again: $$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. So, we got that the result is |d + 3| (the absolute value of d + 3). Absolute value is always non-negative, so the result we got (|d + 3|) cannot be negative, it could be positive or 0.

More about the even roots on the GMAT:

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root. The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
_________________
Intern  B
Joined: 14 Sep 2016
Posts: 13
Solve square root of (d)^2+6d+9?  [#permalink]

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Quote:
Thank you so much for your explanation. I got confused since in the book they have mentioned that if a square root sign is used in the GMAT, only the positive value is considered. e.g.. $$\sqrt{16}$$ then answer should be only 4 and not both 4 and -4.

Quote:

The book is correct.

Check again: $$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. So, we got that the result is |d + 3| (the absolute value of d + 3). Absolute value is always non-negative, so the result we got (|d + 3|) cannot be negative, it could be positive or 0.

More about the even roots on the GMAT:

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root. The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.

So the book is right, When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root..

But if $$\sqrt{9} = 3$$, NOT +3 or -3
then why is $$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. i.e (d+3) and -(d+3). Shouldn't it be d+3 ONLY?

Sorry I am really confused.
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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aj3001 wrote:
But if $$\sqrt{9} = 3$$, NOT +3 or -3
then why is $$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. i.e (d+3) and -(d+3). Shouldn't it be d+3 ONLY?

Sorry I am really confused.

No. Absolute value sign, ||, ensures that the result is not negative: |d + 3| cannot be negative.

If d >=-3, then |d + 3| = d + 3 (notice that if d >= -3, then d + 3 is NOT negative).
If d < -3, then |d + 3| = -(d + 3) (notice that if d < -3, then -(d + 3) is NOT negative).

So, as you can see the result is non-negative in all cases. Try plugging numbers to check.
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Intern  B
Joined: 14 Sep 2016
Posts: 13
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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Bunuel wrote:
aj3001 wrote:
But if $$\sqrt{9} = 3$$, NOT +3 or -3
then why is $$\sqrt{d^2+6d+9}=\sqrt{(d+3)^2}=|d+3|$$. i.e (d+3) and -(d+3). Shouldn't it be d+3 ONLY?

Sorry I am really confused.

No. Absolute value sign, ||, ensures that the result is not negative: |d + 3| cannot be negative.

If d >=-3, then |d + 3| = d + 3 (notice that if d >= -3, then d + 3 is NOT negative).
If d < -3, then |d + 3| = -(d + 3) (notice that if d < -3, then -(d + 3) is NOT negative).

So, as you can see the result is non-negative in all cases. Try plugging numbers to check.

Got it, thank you so much! Intern  B
Joined: 25 Sep 2018
Posts: 7
Re: Solve square root of (d)^2+6d+9?  [#permalink]

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If you take the square roof of a number there are always 2 possible solutions:
E.g.:
(-x)^2 = x^2
x^2 = x^2
Hope this helps

Posted from my mobile device
Manhattan Prep Instructor G
Joined: 04 Dec 2015
Posts: 833
GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: Solve square root of (d)^2+6d+9?  [#permalink]

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Reactzz wrote:
If you take the square roof of a number there are always 2 possible solutions:
E.g.:
(-x)^2 = x^2
x^2 = x^2
Hope this helps

Posted from my mobile device

That's true, but not technically true on the GMAT. A lot of people find this counterintuitive, so here's an article about the square root situation:
https://www.manhattanprep.com/gmat/blog ... -the-gmat/

For the original question, Bunuel's answer is right. -(d+3) isn't necessarily a negative number - for instance, if d = -100, then -(d+3) is positive. So, -(d+3) could actually be the positive square root!
_________________ Chelsey Cooley | Manhattan Prep | Seattle and Online

My latest GMAT blog posts | Suggestions for blog articles are always welcome! Re: Solve square root of (d)^2+6d+9?   [#permalink] 26 Jan 2019, 14:36
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