Write an equation that has the following roots: 2, -1, 5
Answer key: x^3 - 6x^2 + 3x + 10 = 0
For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?
Thanks.
Write an equation that has the following roots: 2, -1, 5
Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?
If we have these roots: 1, 1 + √2, 1 - √2
the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.
When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,
it becomes:
(x^2 -x -x√2 -x +1 +√2) (x -1 +√2)
collect like terms:
(x^2 -2x -x√2 +1 +√2) (x -1 +√2)
To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.
Is there a method to write an equation with 3 given roots (other than the above method)?
Thanks.
You have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply
the factors corresponding to these first you get:
(x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b))
................=x^2 - 2a x + (a^2-b)
Which leaves you with the easier final step of computing:
(x-1)(x^2 - 2a x + (a^2-b))
RonL
Hello, shenton!
The sum and product of roots works well for quadratic equations.
For higher-degree equations, there is a generalization we can use.
To make it simple (for me), I'll explain a fourth-degree equation.
Divide through by the leading coefficient: .$\displaystyle x^4 + Px^3 + Qx^2 + Rx + S \:=\:0$
Insert alternating signs: .$\displaystyle +\:x^4 - Px^3 + Qx^2 - Rx + S \:=\:0$
. . . . . . . . . . . . . . . . . . $\displaystyle \uparrow\quad\;\; \uparrow\qquad\;\;\uparrow\qquad\;\,\uparrow\qquad \,\uparrow$
Suppose the four roots are: $\displaystyle a,\,b,\,c,\,d.$
The sum of the roots (taken one at a time) is: $\displaystyle -P.$
. . $\displaystyle a + b + c + d \:=\:-P$
The sum of the roots (taken two at a time) is: $\displaystyle Q.$
. . $\displaystyle ab + ac + ad + bc + bd + cd \:=\:Q$
The sum of the roots (taken three at a time) is: $\displaystyle -R.$
. . $\displaystyle abc + abd + acd + bcd \:=\:-R$
The sum of the roots ("taken four at a time") is: $\displaystyle S.$
. . $\displaystyle abcd \:=\:S$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For your problem with roots: $\displaystyle (a,b,c) \:=\:(2,-1,5)$
. . we have: .$\displaystyle x^3 + Px^2 + Qx + R \:=\:0$
Then: .$\displaystyle a + b + c \:=\:-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P$
. . Hence: $\displaystyle P = -6$
And: .$\displaystyle ab + bc + ac \:=\:Q\quad\Rightarrow\quad(2)(-1) + (-1)(5) + (2)(5) \:=\:Q$
. . Hence: $\displaystyle Q = 3$
And: .$\displaystyle abc \:=\:-R\quad\Rightarrow\quad(2)(-1)(5)\:=\:-R$
. . Hence: $\displaystyle R = 10$
Therefore, the cubic is: .$\displaystyle x^3 - 6x^2 + 3x + 10 \:=\:0$
This is awesome, Soroban. Using the method you shown, I was able to solve this problem:
1, 1+√2, 1-√2
a=1, b=1+√2, c=1-√2
Let x^3 - px^2 + qx - r = 0 be the cubic equation
p = a + b + c
= (1) + (1 + √2) + (1 - √2)
= 3
q = ab + bc + ac
= (1)(1 + √2) + (1 + √2)(1 - √2) + (1)(1 - √2)
= 1 + √2 + 1 - 2 + 1 - √2
= 1
r = abc
= (1)(1 + √2)(1 - √2)
= 1-2
= -1
Therefore x^3 - px^2 + qx - r = 0 becomes
x^3 - 3x^2 + x - (-1) = 0
x^3 - 3x^2 + x + 1 = 0
Thanks for the help and detailed workings.