If #3x^2-4x+1# has zeros #alpha# and #beta#, then what quadratic has zeros #alpha^2/beta# and #beta^2/alpha# ?

3 Answers
Feb 14, 2018

Find #alpha# and #beta# first.

Explanation:

#3x^2 - 4x + 1 = 0#
The left side factors, so that we have
#(3x - 1)(x - 1) = 0#.
Without loss of generality, the roots are #alpha = 1# and #beta = 1/3#.

#alpha^2/beta = 1^2/(1/3) = 3# and #(1/3)^2/1= 1/9#.

A polynomial with rational coefficients having these roots is
#f(x) = (x - 3)(x - 1/9)#
If we desire integer coefficients, multiply by 9 to obtain:
#g(x) = 9(x - 3)(x - 1/9) = (x - 3)(9x - 1)#
We may multiply this out if we wish:
#g(x) = 9x^2 - 28x + 3#

NOTE: More generally, we might write
#f(x) = (x - alpha^2/beta)(x - beta^2/alpha)#
#= x^2 - ((alpha^3 + beta^3)/(alphabeta))x + alphabeta#

Feb 14, 2018

#9x^2-28x+3#

Explanation:

Note that:

#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta#

and:

#(x-alpha^2/beta)(x-beta^2/alpha) = x^2-(alpha^2/beta+beta^2/alpha)x+(alpha^2/beta)(beta^2/alpha)#

#color(white)((x-alpha^2/beta)(x-beta^2/alpha)) = x^2-(alpha^3+beta^3)/(alpha beta) x + alpha beta#

#color(white)((x-alpha^2/beta)(x-beta^2/alpha)) = x^2-((alpha+beta)^3-3alpha beta(alpha+beta))/(alpha beta) x + alpha beta#

In our example, dividing #3x^2-4x+1# by #3# we have:

#{ (alpha+beta = 4/3), (alpha beta=1/3) :}#

So:

#((alpha+beta)^3-3alpha beta(alpha+beta))/(alpha beta) = ((4/3)^3-3(1/3)(4/3))/(1/3) = (64/27-4/3)/(1/3) = 28/9#

So the desired polynomial can be written:

#x^2-28/9x+1/3#

Multiply through by #9# to get integer coefficients:

#9x^2-28x+3#

Feb 14, 2018

Proposed solution below;

Explanation:

#3x²-4x+1#

Note: #a# is alpha, #b# is beta

#a + b = 4/3#

#ab =1/3#

To form an equation we find the sum and products of the roots..

For Sum

#(a²)/b + (b²)/a = (a^3 + b^3)/(ab)#

But; #a^3 + b^3 = (a+b)³-3ab(a+b)#

Therefore;

#((a+b)³-3ab(a+b))/(ab)#

Hence we substitute the values..

#((4/3)³-3(1/3)(4/3))/(1/3)#

#((64/27)-cancel3(1/cancel3)(4/3))/(1/3)#

#(64/27 - 4/3)/(1/3)#

#((64 - 36)/27)/(1/3)#

#(28/27)/(1/3)#

#(28/27) div (1/3)#

#(28/27) xx (3/1)#

#(28/cancel27_9) xx (cancel3/1)#

#28/9#

Hence, the sum is #28/9#

For Products

#((a²)/b)((b²)/a)#

#((ab)²)/(ab)#

#(1/3)^2 div 1/3#

#1/9 div 1/3#

#1/9 xx 3/1#

#1/cancel9_3 xx cancel3/1#

#1/3 xx 1/1#

#1/3#

Hence, the product is #1/3#

#x²-(a+b)x+ab#

#x²-(28/9)x+1/3#

#9x²-28x+3#

Multiplying through by #9#

Hope this helps!