Asked by echo
if alpha and beta are the zeros of the polynomial 2x^2-4x+5 then find the values of
(i)alpha^2+beta^2
(ii)1/alpha^2+1/beta^2
(iii)(r)alpha/beta+(r)beta/alpha
(iv)alpha^-1+beta^-1
(i)alpha^2+beta^2
(ii)1/alpha^2+1/beta^2
(iii)(r)alpha/beta+(r)beta/alpha
(iv)alpha^-1+beta^-1
Answers
Answered by
MathMate
The general solution to the equation
ax^2+bx+c=0
is, using the quadratic formula:
α=(-b+√(b²-4ac))/2a, and
β=(-b-√(b²-4ac))/2a
From there, expand α²+β² to get (b²-2ac)/a²
I will leave parts (ii) to (iv) for you as practice.
ax^2+bx+c=0
is, using the quadratic formula:
α=(-b+√(b²-4ac))/2a, and
β=(-b-√(b²-4ac))/2a
From there, expand α²+β² to get (b²-2ac)/a²
I will leave parts (ii) to (iv) for you as practice.
Answered by
mathew
alpha+beta=-b/a
=--4/2
=2
alpha^2+beta^2=(alpha+beta)^2
=2^2
=4
(ii)take reciprocal(1/2)^2=1/4
(iv)a^-1=1/a similarly 1/2
tried my best to answer hope its correct:)
your pal
mathew
=--4/2
=2
alpha^2+beta^2=(alpha+beta)^2
=2^2
=4
(ii)take reciprocal(1/2)^2=1/4
(iv)a^-1=1/a similarly 1/2
tried my best to answer hope its correct:)
your pal
mathew
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