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555-605 (Medium)|   Algebra|   Roots|      
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If g(x) = 3x + sqrt x Updated on: 25 Feb 2016, 17:08
Originally posted by jwam2 on 25 Feb 2016, 15:31.
Last edited by ENGRTOMBA2018 on 25 Feb 2016, 17:08, edited 1 time in total.
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If \(g(x) = 3x + \sqrt {x}\), what is the value of \(g(d^2 +6d+9)\)?



(Chapter 5, Page 79 - In Action problem #2)

SEO was poor for this question. Also, this is more open ended practice so it's less GMAT like.
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If g(x) = 3x + sqrt x 25 Feb 2016, 19:23
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jwamala
If \(g(x) = 3x + \sqrt {x}\), what is the value of \(g(d^2 +6d+9)\)?



(Chapter 5, Page 79 - In Action problem #2)

SEO was poor for this question. Also, this is more open ended practice so it's less GMAT like.
Show more

Hi,
1)\(g(x) = 3x + \sqrt {x}\)..
and \(g(d^2 +6d+9)\)..
so substitute x as d^2 +6d+9..
\(g(d^2 +6d+9) = 3(d^2 +6d+9) + \sqrt {d^2 +6d+9}\)..
\(3d^2+18d+27+d+3=3d^2+19d+30\)

or
2)\(3x + \sqrt {x}=\sqrt {x}(3\sqrt {x}+1)\)...
if\(x= d^2 +6d+9\), \(\sqrt{x}\)=\(\sqrt{(d+3)^2}\)..
\(\sqrt {x}(3\sqrt {x}+1)\)=\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)..
\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)...
=\((d+3)(3d+9+1)=(d+3)(3d+10)\)
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If g(x) = 3x + sqrt x 25 Feb 2016, 19:54
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The solutions suggest that "3d^2 + 17d + 24" is also an answer. I'm not fully convinced so I've posted it here.

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If g(x) = 3x + sqrt x 25 Feb 2016, 20:15
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I see but still not convinced on having a negative root.

Source: MGMAT Chapter 5, Page 79 - In Action problem #2)

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If g(x) = 3x + sqrt x 25 Feb 2016, 20:19
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jwamala
I see but still not convinced on having a negative root.

Source: MGMAT Chapter 5, Page 79 - In Action problem #2)

Posted from my mobile device
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As per GMAT,
\(\sqrt{9}= 3\) and not -3..
only if it is given \(\sqrt{9}= |3|\), you take two values 3 and -3..

there must be some thing else given too to take -ive root as an option..
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If g(x) = 3x + sqrt x 25 Feb 2016, 20:23
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I agree with you; that's why I posted the question.

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If g(x) = 3x + sqrt x 18 Aug 2017, 04:10
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chetan2u
jwamala
I see but still not convinced on having a negative root.

Source: MGMAT Chapter 5, Page 79 - In Action problem #2)

Posted from my mobile device

As per GMAT,
\(\sqrt{9}= 3\) and not -3..
only if it is given \(\sqrt{9}= |3|\), you take two values 3 and -3..

there must be some thing else given too to take -ive root as an option..
Show more


Is this only as per GMAT or mathematics in General? bunuel?
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If g(x) = 3x + sqrt x 18 Aug 2017, 04:19
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chetan2u
jwamala
I see but still not convinced on having a negative root.

Source: MGMAT Chapter 5, Page 79 - In Action problem #2)

Posted from my mobile device

As per GMAT,
\(\sqrt{9}= 3\) and not -3..
only if it is given \(\sqrt{9}= |3|\), you take two values 3 and -3..

there must be some thing else given too to take -ive root as an option..


Is this only as per GMAT or mathematics in General? bunuel?
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When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. Even roots have only a positive value on the GMAT.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
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If g(x) = 3x + sqrt x 01 Mar 2025, 18:06
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(d+9)^2 + (d+3) ???
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If g(x) = 3x + sqrt x 02 Mar 2025, 03:08
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Please someone correct me if i am mistaken, but you are not allowed to distribute the square root to a binomial unless you solve the binomial and then perform the square root. I can't see the answer choices but surely the quagratic equals d = -3 hence we can solve for g(-3).
chetan2u
jwamala
If \(g(x) = 3x + \sqrt {x}\), what is the value of \(g(d^2 +6d+9)\)?



(Chapter 5, Page 79 - In Action problem #2)

SEO was poor for this question. Also, this is more open ended practice so it's less GMAT like.

Hi,
1)\(g(x) = 3x + \sqrt {x}\)..
and \(g(d^2 +6d+9)\)..
so substitute x as d^2 +6d+9..
\(g(d^2 +6d+9) = 3(d^2 +6d+9) + \sqrt {d^2 +6d+9}\)..
\(3d^2+18d+27+d+3=3d^2+19d+30\)

or
2)\(3x + \sqrt {x}=\sqrt {x}(3\sqrt {x}+1)\)...
if\(x= d^2 +6d+9\), \(\sqrt{x}\)=\(\sqrt{(d+3)^2}\)..
\(\sqrt {x}(3\sqrt {x}+1)\)=\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)..
\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)...
=\((d+3)(3d+9+1)=(d+3)(3d+10)\)
Show more
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If g(x) = 3x + sqrt x 02 Mar 2025, 04:17
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foookato
Please someone correct me if i am mistaken, but you are not allowed to distribute the square root to a binomial unless you solve the binomial and then perform the square root. I can't see the answer choices but surely the quagratic equals d = -3 hence we can solve for g(-3).
chetan2u
jwamala
If \(g(x) = 3x + \sqrt {x}\), what is the value of \(g(d^2 +6d+9)\)?



(Chapter 5, Page 79 - In Action problem #2)

SEO was poor for this question. Also, this is more open ended practice so it's less GMAT like.

Hi,
1)\(g(x) = 3x + \sqrt {x}\)..
and \(g(d^2 +6d+9)\)..
so substitute x as d^2 +6d+9..
\(g(d^2 +6d+9) = 3(d^2 +6d+9) + \sqrt {d^2 +6d+9}\)..
\(3d^2+18d+27+d+3=3d^2+19d+30\)

or
2)\(3x + \sqrt {x}=\sqrt {x}(3\sqrt {x}+1)\)...
if\(x= d^2 +6d+9\), \(\sqrt{x}\)=\(\sqrt{(d+3)^2}\)..
\(\sqrt {x}(3\sqrt {x}+1)\)=\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)..
\(\sqrt{(d+3)^2}(3\sqrt{(d+3)^2}+1)\)...
=\((d+3)(3d+9+1)=(d+3)(3d+10)\)
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There is no where mentioned that \(d^2 +6d+9\) is equal to zero so you cannot conclude that d = -3

You have to take \(x = (d + 3)^2\) and solve \(g((d + 3)^2)\)
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