Question

Find the equation whose roots are given by adding 2 to the roots of the equation

\[\large\rm x^4+3x^3-13x^2-51x-36=0 \ \]




Note:- We are not allowed to first solve the equation. The second equation should be given without solving the equation

Example
second coefficient = b/a = -Σa = -3
Since we are adding two to every root so
b/a = -(Σa +2) = -(Σa +Σ2) = -(-3+8) = -5
thus the equation should have x^4-5x³

Answer 1

@Evoker @ganeshie8

Answer 2

hmm i think you could just plug in (x-2) for x in the original equation, essentially the f of (x-2)

Answer 3

basically a horizontal shift.

Answer 4

No that method is also not allowed.
See this question is of 5 marks, in which I also have to find roots of this equation (second part) and find the equation of (root+2). If I used your method, the simplifying of the expression would alone take 5-7 minutes and 3 minutes wont be enough to find the solution of this equation

Answer 5

hmm so looks like we need to look at the general form (x+a)(x+b)(x+c)(x+d) where 2 is added to each term.

Answer 6

yes

Answer 7

before adding 2 yields (x^2+(a+b)x+ab)(x^2+(c+d)x+cd)

Answer 8

or lets see x^4+(c+d)x^3+cdx^2+(a+b)x^3+(a+b)(c+d)x^2+(a+b)(cd)x+abx^2+ab(c+d)x+abcd

Answer 9

which simplifies to x^4+(a+b+c+d)x^3+(ab+cd+(a+b)(c+d))x^2+((a+b)(cd)+ab(c+d))x+abcd

Answer 10

in other words
x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

Answer 11

thanks didn't have that to look at

Answer 12

Have to go for now.

Answer 13

ok so sum of (a+2)(b+2) would equal sum of ab +2 sum of a + 2 sum of b + sum of 4

Answer 14

Yes that the part I am having trouble with
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16
which amounts to -9
Correct coefficient is -7

Answer 15

Oh I am pretty sure it has to be (a-2)(b-2)

Answer 16

Ah that may be the problem.

Answer 17

But the coefficient is again wrong

Answer 18

It doesn't take that long to sub in \(x-2\) especially if you know the binomial theorem.

Answer 19

Yes, I think you're meant to sub in \(x-2\), it turns out "nicely".

Answer 20

Its not about lengthiness, I am not scared to do all the work. Its just that I want to know what mistake I am doing when I am applying the method

Answer 21

I did the line your having trouble with by hand and i get sum of ab + 6a + 6*4, the problem is the index is different for the two types of sums.

Answer 22

sum of i versus some of i,j where i does not equal j.

Answer 23

Well this bit is wrong already...

(quote)Yes that the part I am having trouble with
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16
which amounts to -9
Correct coefficient is -7(quote)

Σab is representative of sum of roots taken two at a time
This is nominally, for four roots (remember this is a quartic!), say alpha, beta, gamma and delta
\[\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta\]
Thus you are required to find out how the increase in roots affects ALL of these terms.

So this exercise is actually much trickier than replace x with alpha + 2, since P(alpha) is 0 in the new expression, then P(x-2) will give 0 for given alphas (roots). I would not recommend this coefficient approach which is actually more complicated in the long run.

You do use coefficients if asked to evaluate expression for roots such as
\[\alpha^4 + \beta ^4 + \gamma^4 + \delta^4\] where it is advisable to use these rather than evaluate the roots themselves.

Answer 24

@FaiqRaees. See my response above.

Answer 25

@ganeshie8

Answer 26

@mww How should I use them when they get much complex (multiples of each other). Wouldn't it become much tiring to calculate each and every multiple and then add them and hence increase the chance to make an error. And the method I have adopted is a certified, universal method not a method which I came up with myself
@Evoker How did you get sum of ab + 6a + 6*4

Answer 27

@jim_thompson5910

Answer 28

@FaiqRaees
I think @Evoker's method is still by far the best.

The original equation is
\(x^4+3x^3-13x^2-51x-36\)
Let's say it takes you 5 minutes to do the expansion and get
\(x^4-5x^3-7x^2+5x+6\).................(answer to part (a))
Compare with the original, and apply Descarte's rule to both, you will find that there is an extra positive root, which means that +1 is a root to the new equation. +2 is ruled out because the original equation does not have zero as a root.

While we're there, if we invert the sign of the odd degree terms, we see that the sum of coefficients is still zero, hence x=-1 is also a root.

Divide the original equation by \((x^2-1)\) gives \(x^2-5x-6\) which easily factors into \((x-6)(x+1)\), giving the four roots as
S={-1(mult. 2), +1,+6} ...................(answer to part (b))

The roots of the original equation is S with each root reduced by 2. That should complete the answer to part B.

Answer 29

I took (a+2)(b+2) + (a+2)(c+2)+...... when you do this you do get 4 choose 2 or 6 equations each having a coefficient of 4, also each has a 2a + 2b with 3 of the 6 equations having each variable so get 6 times the sum of a, That is where I got the final result of sum of ab + 6 sum of a + 6*4.

Answer 30

@mathmate

\[x^4-5x^3-7x^2+5x+6= \]
\[(x^4-5x^3-6x^2)-(x^2+5x-6)=\]
\[x^2(x^2+5x-6)-(x^2+5x-6)=\]
\[(x^2-1)(x^2+5x-6)=\]
\[(x-1)(x+1)(x-2)(x-3)\]

Answer 31

ooops, i made a sign error with the \(5x\) term, but hopefully you get the point :P

Answer 32

http://math.stackexchange.com/questions/1876707/forming-a-quartic-equation-by-given-roots/1876714?noredirect=1#comment3847470_1876714
http://math.stackexchange.com/questions/1876857/specific-double-summation-problems/1876862?noredirect=1#comment3847541_1876862

These were what I was talking about