Polynomial Question (1 Viewer)

[Georgegeorge]

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Write down the equation whose roots are the squares of the roots of:
2x² + 3x + 5 = 0

and more specifically, how do you get to
x² - (a² + B²)x + a²B² = 0, knowing that the required equation must have roots a² and B² with the roots obviously being a and B

Is it the same process for: find the equation whose roots are twice the equation
3x³ - 2x² + 1 = 0 and

Find the cubic equation whose roots are squares of that of
x³ + 2x +1 = 0

I anybody replies to this, thankyou for your time!!!!

 

[Mattamz]

Is it the same process for: find the equation whose roots are twice the equation
3x³ - 2x² + 1 = 0

let x = x/2

3(x/2)^3 - 2(x/2)^2 + 1 = 0
3x^3/8 - x/2 + 1 = 0
3x^3 - 4x + 8 = 0


Find the cubic equation whose roots are squares of that of
x³ + 2x +1 = 0

let x = x^(1/2)

x^(3/2) +2x^(1/2) = -1
x^(1/2)[x+2] = -1
x[x+2]^2 = 1
x(x^2 +4x + 4) = 1
x^3 + 4x^2 +4x - 1 = 0

 

[ssglain]

Strictly speaking, this type of questions is of the Ext-2 course, but I'm not sure to what extent it is included in the Ext-1 syllabus.

Taking this as an example: find the equation whose roots are twice those of P(x) = 3x³ - 2x² + 1 = 0

The concept is very simple:
Let the roots of P(x) = 0 be a, b, c and let y = 2x.
Clearly, if y satisfies P(y) = (y - A)(y - B)(y - C) = 0, the roots A, B, C are in fact 2a, 2b, 2c.

Substituting x = y/2 into 3x³ - 2x² + 1 = 0:
3(y³/8) - 2(y²/4) + 1 = 0
.: P(y) = 3y³ - 4y² + 8 = 0

Some teachers, such as mine, will require you to change all the y variables back to x but it doesn't really matter.

 

[Georgegeorge]

Thank-you everbody