Question

Forming New Quadratic Equations HELP!!!

Answer 1

The roots of the equation x^2-5x+2=0 are a and b. find the new quadratic equation whose roots will me a^3 and b^3

Answer 2

can you guys do it by letting m=a^3? thanks

Answer 3

ive gotten to m^2-125m+8=30m^(1/3) (2-m) but now im stuck

Answer 4

For a quadratic equation, x^2 - (sum of roots) x + (product of roots) = 0
Consider x^2-5x+2=0
sum of roots = a+b = 5
product of roots = ab = 2
a^2 + b^2 = (a+b)^2 - 2ab = 5^2 - 2(2) = ...

Consider the new equation with roots a^3 and b^3
Sum of roots = a^3 + b^3 = (a+b)(a^2-ab+b^2) = ...
Product of roots = a^3b^3 = (ab)^3 = ...
Plug the above into x^2 - (sum of roots) x + (product of roots) = 0 to get the new equation.

Answer 5

can you do it by letting m=a^3?

Answer 6

May I know what m is?

Answer 7

m is the root for the new equation

Answer 8

and you sub the cube root of m into the original equation

Answer 9

So, you get m^2-5m+2=0?

Answer 10

nope replace m with the cube root of m and then you have to get rid of the cube roots to get the new equation

Answer 11

you can use that method if the roots are **multiples** of given equation. it wont be that useful when u want to find an equation whose roots are squares of cubes of given roots

Answer 12

it works with the squares of roots though

Answer 13

it wont be simple, cuz u wud endup with a radical for the x coeffecient

Answer 14

right

Answer 15

i like the Callisto method for squares and cubes.... its simple and elegant

Answer 16

\[(\sqrt[3]{m})^2-5(\sqrt[3]{m})+2=0\]\[(\sqrt[3]{m}-\frac{5}{2})^2-\frac{17}{4}=0\]\[(\sqrt[3]{m}-\frac{5}{2})^2-\frac{17}{4}=0\]Ok, you solve m from here. Then, you get the two roots for m, and you can form a new equation using the roots of m. But you just make it complicated then.

Answer 17

ahhh i see, ill stick with your method for cubics then. Thanks :D

Answer 18

Welcome :)