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Third Degree Equations

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Third Degree Equations [#permalink]

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Here is a post on solving third degree equations:

How do you solve an equation such as \(x^3 - 6x^2 + 11x - 6 = 0\)?
How do we get the values of x which satisfy this equation?

If you do get a third degree equation, it will have one very easy root such as 0 or 1 or -1 or 2 or -2 etc. Try a few of these values to get the first root. Here x = 1 works. It is easy to see since you have two 6s, a 1 and an 11 as the coefficients.
Putting x = 1: \((1)^3 - 6(1)^2 + 11(1) - 6 = 0\)
So you know that (x – 1) is a factor. Now figure out the quadratic which when multiplied by (x – 1) gives the third degree expression

\((x - 1)(ax^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)
How do we figure out the values of a, b and c? Let’s see.
Coefficient of x^3 on right hand side is 1. So you know that all you need is x^2 so that it multiplies with x to give x^3 on left hand side too. So ‘a’ must be 1.

\((x - 1)(x^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)
The constant term, ‘c’, is easy to figure out too. It should multiply with -1 to give -6 on right hand side. Hence c must be 6.

\((x - 1)(x^2 + bx + 6) = x^3 - 6x^2 + 11x - 6\)
Getting the middle term is slightly more complicated. bx multiplies with x to give x^2 term and you also get the x^2 term by multiplying -1 with x^2. You have -6x^2 on right hand side so you need the same on the left hand side too. You already have -x^2 (by multiplying -1 with x^2) so you need another -5x^2 from bx^2. So b must be -5.

\((x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6\)
Now you just factorize the quadratic in the usual way. Let’s see how exactly we would do it using a question.

Question: Is \(x^3 + 2x^2 - 5x - 6 < 0\)
Statement 1: -3 < x <= -1
Statement 2: -1 <= x < 2

Solution: We know how to deal with inequalities with multiple factors (discussed http://www.veritasprep.com/blog/2012/06 ... e-factors/).
But the inequality is not split into factors here so we will have to do it on our own.
Let’s first try to find the simple root that this expression must have. Try x = 1, -1 etc. We see that when we put x = -1, the expression becomes 0.
\(x^3 + 2x^2 - 5x - 6 = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = 0\)
So the first factor we get is (x + 1).

\((x + 1)(ax^2 + bx + c) = x^3 + 2x^2 - 5x - 6\)
a must be 1 since we have \(x^3\) on the right hand side.
c must be -6 since we have -6 on the right hand side.

\((x + 1)(x^2 + bx - 6) = x^3 + 2x^2 - 5x - 6\)
\(bx^2 + x^2 = 2x^2\)
So b must be 1.

We get: \((x + 1)(x^2 + x - 6)\) which is equal to \((x + 1)(x + 3)(x - 2)\) (after splitting the quadratic)
So the question becomes:
Is \((x + 1)(x + 3)(x - 2) < 0?\)
We already know how to deal with inequalities with multiple factors. The transition points here will be -3, -1 and 2. The expression will be negative in the ranges -1 < x < 2 and x < -3

Statement 1: -3 < x <= -1
In this region, the expression is positive (when -3 < x < -1) or 0 (when x = -1). Hence it will certainly not be negative. This is sufficient to answer the question with ‘No’. Hence statement 1 is sufficient to answer the question.

Statement 2: -1 <= x < 2
In this region, the expression is negative (when -1 < x < 2) or 0 (when x = -1). We cannot say for certain whether it will be negative or not. Hence statement 2 alone is not sufficient to answer the question.

Answer (A)
It’s not hard to deal with third degree equations. All you have to do is bring it down to second degree by figuring out one root and then the problem is in a format you already know.
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Re: Third Degree Equations [#permalink]

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New post 13 Sep 2013, 00:35
Is it possible to use plug-in approach to quickly solve similar questions?
Statement 1: -3 < x <= -1
Plug-in: -3,-2,-1
Statement 2: -1 <= x < 2
Plug-in: -1,0,1,2

I understand the range is important in such questions, but then ain't there methods to plug-in and reverse solve?
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Re: Third Degree Equations [#permalink]

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New post 13 Sep 2013, 03:58
Disillusioned wrote:
Is it possible to use plug-in approach to quickly solve similar questions?
Statement 1: -3 < x <= -1
Plug-in: -3,-2,-1
Statement 2: -1 <= x < 2
Plug-in: -1,0,1,2

I understand the range is important in such questions, but then ain't there methods to plug-in and reverse solve?


Sure you can but there will be risk since you certainly cannot plug in all values to check.

e.g. Imagine that you have a third degree expression which factors out as:
Is (x + 1)(x + 2)(x - 3) <= 0?

Say, given range is -2 < x < 3
You try -1, 0, 1, 2
All of them satisfy the inequality. Does it mean that in this range, (x + 1)(x + 2)(x - 3) is negative or 0? Does it mean that this statement alone is sufficient?
No. because in -2 < x< -1, (x + 1)(x + 2)(x - 3) is positive. While in -1 < x < 3, the expression is negative.

So I like to use plug in only when I don't see any other easy way out.

As for questions which use plug in to reverse solve, the higher level questions nowadays do not allow that luxury. They are made keeping in mind that you know how to plug in. Only the lower level questions might be solvable as such.
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Re: Third Degree Equations [#permalink]

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New post 25 Oct 2016, 03:45
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Re: Third Degree Equations   [#permalink] 25 Oct 2016, 03:45
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