If #3x^2-4x+1# has zeros #alpha# and #beta#, then what quadratic has zeros #alpha^2/beta# and #beta^2/alpha# ?
3 Answers
Find
Explanation:
The left side factors, so that we have
Without loss of generality, the roots are
A polynomial with rational coefficients having these roots is
If we desire integer coefficients, multiply by 9 to obtain:
We may multiply this out if we wish:
NOTE: More generally, we might write
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
#{ (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
#9x^2-28x+3#
Proposed solution below;
Explanation:
Note:
To form an equation we find the sum and products of the roots..
For Sum
But;
Therefore;
Hence we substitute the values..
Hence, the sum is
For Products
Hence, the product is
Multiplying through by
Hope this helps!