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How many roots does x^6 –12x^4 + 32x^2 = 0 have?

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Bunuel
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How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   10 Mar 2016, 05:10
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E

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62% (01:29) correct 38% (01:47) wrong based on 148 sessions
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How many roots does x^6 –12x^4 + 32x^2 = 0 have?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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E
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davedekoos
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How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   Updated on: 29 Mar 2017, 08:58
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To find the roots we can factor the polynomial down to first order expressions and count how many roots we get.

The first thing to do is to factor out \(x^2\)

Then we have \(x^2(x^4-12x^2+32)=0\)

Now to make things more familiar we can substitute \(y=x^2\)

Now it looks like this
\(y(y^2-12y+32)=0\)

And we can factor it like we are used to

\(y(y-4)(y-8)=0\)

y=0
y=4
y=8

So substituting \(x^2\) back in for y:

\(x^2=0\)
\(x^2=4\)
\(x^2=8\)

So then the roots are

\(x=0\)
\(x=-2\)
\(x=2\)
\(x=-\sqrt{8}\)
\(x=\sqrt{8}\)

5 roots. Answer: E

Note, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the \(x^2\), we have \(x^2\) and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the \(x^2\). Total roots = 5.

Originally posted by davedekoos on 10 Mar 2016, 14:28.
Last edited by davedekoos on 29 Mar 2017, 08:58, edited 1 time in total.
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How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   21 Mar 2017, 19:59
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Hi mesutthefail,

The GMAT often tests you on rules/patterns that you know, but sometimes in ways that you're not used to thinking about. This prompt is really just about factoring and Classic Quadratics, but it looks a lot more complicated than it actually is.

A big part of properly dealing with a Quant question on the GMAT is in how you organize the information and 'simplify' what you've been given. This prompt starts us off with....

X^6 –12X^4 + 32X^2 = 0

This certainly looks complex, but if you think about how you can simplify it, then you'll recognize that can 'factor out' X^2 from each term. This gives us...

(X^2)(X^4 - 12X^2 + 32) = 0

Now we have something a bit more manageable. While you're probably used to thinking of Quadratics such as X^2 + 6X + 5 as (X+1)(X+5), that same pattern exists here - it's just the exponents are slightly different (even though the math rules are exactly the SAME). We can further rewrite the above equation as...

(X^2)(X^2 -4)(X^2 - 8) = 0

At this point, you don't really need to calculate much, since each 'piece' of the product should remind you of a pattern that you already know.....

X^2 = 0 --> 1 solution
(X^2 - 4) = 0 --> 2 solutions
(X^2 - 8) = 0 --> 2 solutions

Final Answer:
Show Spoiler
E


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mesutthefail
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Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   21 Mar 2017, 02:27
I thought polynoms was not a subject in GMAT ? This questions contains quite the polynomial solutions.
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mesutthefail
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Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   21 Mar 2017, 02:31
davedekoos wrote:
To find the roots we can factor the polynomial down to first order expressions and count how many roots we get.

The first thing to do is to factor out \(x^2\)

Then we have \(x^2(x^4-12x^2+32)=0\)

Now to make things more familiar we can substitute \(y=x^2\)

Now it looks like this
\(y(y^2-12y+32)=0\)

And we can factor it like we are used to

\(y(y-4)(y-8)=0\)

y=0
y=4
y=8

So putting it back to x^2

\(x^2=0\)
\(x^2=4\)
\(x^2=8\)

So then the roots are

\(x=0\)
\(x=-2\)
\(x=2\)
\(x=-\sqrt{8}\)
\(x=\sqrt{8}\)

5 roots. Answer: E

Note, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the \(x^2\), we have \(x^2\) and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the \(x^2\). Total roots = 5.


A question. When you first factored out by x^2, the x expression in 32x^2 disappeared completely, however when you second factored out the "y" expression, 12y^2 became 12y instead of 12. Can you please clarify?
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How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink] New post   06 Jun 2020, 23:37
Bunuel wrote:
How many roots does x^6 –12x^4 + 32x^2 = 0 have?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


Asked: How many roots does x^6 –12x^4 + 32x^2 = 0 have?

Let \(x^2 = t\)

\(t^3 - 12t^2 + 32t = 0\)
t(t-8)(t-4) = 0
t = {0,4,8}

\(x =\{0,-2,2,\sqrt{8}, - \sqrt{8}\}\); 5 roots

IMO E
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How many roots does x^6 –12x^4 + 32x^2 = 0 have? [#permalink]
06 Jun 2020, 23:37
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