Question

If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1/α3 + 1/β3 – 2(α 2 + β 2

Answer 1

Answer:

MATHS

If α and β are zeroes of the quadratic polynomial f(x)=3x2−5x−2, then find the values of βα2+αβ2

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ANSWER

f(x)=3x2−5x−2

α+β=35,αβ=3−2

Given, βα2+αβ2

=αβα3+β3

=αβ(α+β)(α2+β2−αβ)

=αβ(α+β)∫(α+β)2−3αβ

=−2/3(35)(925+<

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

a = 1, b = -5, c = 4

sum of roots, α + β = -b/a = 5

Product of roots, αβ = c/a = 4.

1/α + 1/β = α + β / αβ = 5/4

Cubing on both sides

(1/α + 1/β)³ = (5/4)³

We know that (a + b)³ = a³ + b³ + 3ab(a + b)

1/α³ + 1/β³ + 3(1/α)(1/β) (1/α + 1/β) = (5/4)³

1/α³ + 1/β³ + 3/4 * 5/4 = (5/4)³

1/α³ + 1/β³ = (5/4)³ - (3/4*5/4) = 5/4 [(5/4)² - 3/4]

                           5/4(25/16 - 3/4) = 5/4 * 13/16 = 65/64 -- (1)

α² + β² = (α + β)² - 2αβ = (5)² - 2 * 4 = 17.

Thus 1/α³ + 1/β³ - 2(α² + β²) = 65/64 - 34 = - 2111/34.