if alpha and beta are zeros of polynomial f(x)=px2+qx+r then find (i)alpha2+beta2 (ii)alpha/beta+beta/alpha (iii)alpha3+beta3
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Answers
EXPLANATION.
α, β are the zeroes of the polynomial.
⇒ f(x) = px² + qx + r.
As we know that,
Sum of the zeroes of the quadratic equation.
⇒ α + β = -b/a.
⇒ α + β = -q/p. - - - - - (1).
Products of the zeroes of the quadratic equation.
⇒ αβ = c/a.
⇒ αβ = r/p. - - - - - (2).
To find :
(1) = α² + β².
As we know that,
Formula of :
⇒ (x² + y²) = (x + y)² - 2xy.
Using this formula in the equation, we get.
⇒ (α² + β²) = (α + β)² - 2αβ.
Put the value in the equation, we get.
⇒ (α² + β²) = (-q/p)² - 2(r/p).
⇒ (α² + β²) = q²/p² - 2r/p.
⇒ (α² + β²) = q² - 2pr/p².
(2) = α/β + β/α.
⇒ α/β + β/α = α² + β²/αβ.
⇒ α/β + β/α = [α + β]² - 2αβ/αβ.
⇒ α/β + β/α = [-q/p]² - 2(r/p)/(r/p).
⇒ α/β + β/α = [q²/p² - 2r/p]/(r/p).
⇒ α/β + β/α = [q² - 2pr/p²]/(r/p).
⇒ α/β + β/α = q² - 2pr/p² x p/r.
⇒ α/β + β/α = q² - 2pr/pr.
(3) = α³ + β³.
As we know that,
Formula of :
⇒ (x³ + y³) = (x + y)(x² + y² - xy).
⇒ (x³ + y³) = (x + y)[(x + y)² - 2xy - xy].
⇒ (x³ + y³) = (x + y)[(x + y)² - 3xy].
Using this formula in the equation, we get.
⇒ α³ + β³ = (α + β)[(α + β)² - 3αβ].
Put the value in the equation, we get.
⇒ α³ + β³ = (-q/p)[(-q/p)² - 3(r/p)].
⇒ α³ + β³ = (-q/p)[q²/p² - 3r/p].
⇒ α³ + β³ = (-q/p)[q² - 3pr/p²].
⇒ α³ + β³ = [-q³ + 3pqr]/p³.
⇒ α³ + β³ = 3pqr - q³/p³.
Given :-
f(x) = px² + qx + r
To Find :-
α² + β²
α/β + β/α
α³ + β³
Solution :-
Sum of zeroes = -b/a
Here
b = q
a = p
Sum = -(q)/p
Sum = -p/q
Now
Product of zeroes = c/a
Here
c = r
a = q
Product = r/q
1] α² + β²
(a² + b²) = (a + b)² - 2ab
(α² + β²) = (-q/p)² - 2(r/p)
(α² + β²) = (q²/p²) - 2 × r/p
(α² + β²) = q²/p² - 2r/p
(α² + β²) = q² - 2qr/p²
ii] α/β + β/α
α² + β²/αβ
(a² + b²) = (a + b)² - 2ab
(α + β)² - 2αβ/αβ
(-q/p)² - 2(r/p)/(r/p)
q²/p² - 2r/p/r/p
q² - 2pr/pr
iii]
(α³ + β³) = (α + β){(α + β)² - 3αβ}
(-q/p){(-q/p)² - 3(r/p)}
(-p/q) × {(q²/p²) - 3 × r/p)}
(-p/q) × {(q²/p²) - 3r/p)}
-p/q × q² - 3pr/p²
3pqr - q³/p³