9. If alfa and Bita are the zeroes of polynomial
p (x) = x2 - 7x + 10, find
find the
the quadratic
polynomial with zeroes (-alfa) and (-bita).
please don't give unusefull message
Answers
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Given :
If α & β are the zeroes of the polynomial, x² - 7x + 10 = 0 , then
Find α³ + β³
Solution :
We know that,
⇒ (a³ + b³) = (a + b) (a² - ab + b²)
_
We also know that,
For a quadratic equation of the form :
ax² + bx + c = 0,.
Here, a = 1 , b = -7 , c = 10
Sum of the zeroes = α + β = \frac{-b}{a}
a
−b
Product of the zeroes = αβ = \frac{c}{a}
a
c
⇒ (a³ + b³) = (a + b) (a² - ab + b²)
= (a + b) (a² + 2ab - 3ab + b²)
= (a + b) [ (a² + 2ab + b²) - 3ab ]
⇒ (a + b) [(a + b)² - 3ab]
Hence,
⇒ α³ + β³ = ( \frac{-b}{a} )[(\frac{-b}{a} )^2 - 3(\frac{c}{a})](
a
−b
)[(
a
−b
)
2
−3(
a
c
)]
⇒(\frac{-(-7)}{1})[(\frac{-(-7)}{1})^2 - 3(\frac{10}{1})] = (7)[(7)^2 - 3(10)](
1
−(−7)
)[(
1
−(−7)
)
2
−3(
1
10
)]=(7)[(7)
2
−3(10)]
⇒7(49 - 30) = 7 × 19 = 133
∴ α³ + β³ = 133
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