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

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Math Expert
Joined: 02 Sep 2009
Posts: 49303
How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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10 Mar 2016, 05:10
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Difficulty:

45% (medium)

Question Stats:

56% (01:02) correct 44% (01:23) wrong based on 97 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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How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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21 Mar 2017, 19:59
3
2
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

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Rich
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Joined: 09 Jul 2013
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How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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Updated on: 29 Mar 2017, 08:58
2
4
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

$$x=0$$
$$x=-2$$
$$x=2$$
$$x=-\sqrt{8}$$
$$x=\sqrt{8}$$

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.
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Dave de Koos

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.
##### General Discussion
Intern
Joined: 12 Dec 2016
Posts: 10
Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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21 Mar 2017, 02:27
I thought polynoms was not a subject in GMAT ? This questions contains quite the polynomial solutions.
Intern
Joined: 12 Dec 2016
Posts: 10
Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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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

$$x=0$$
$$x=-2$$
$$x=2$$
$$x=-\sqrt{8}$$
$$x=\sqrt{8}$$

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?
Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have? &nbs [#permalink] 21 Mar 2017, 02:31
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# How many roots does x^6 –12x^4 + 32x^2 = 0 have?

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