α and β are the zeroes of polynomial x 2 + 8x + 6 then find the polynomial whose zeroes are α 2 and β 2
Answers
EXPLANATION.
α and β are the zeroes of the polynomial.
⇒ x² + 8x + 6.
As we know that,
Sum of the zeroes of the quadratic polynomial.
⇒ α + β = - b/a.
⇒ α + β = -(8)/1 = - 8.
Products of the zeroes of the quadratic polynomial.
⇒ αβ = c/a.
⇒ αβ = (6)/1 = 6.
Zeroes of the polynomial.
⇒ α² and β².
Sum of the zeroes of the quadratic polynomial.
⇒ α + β = - b/a.
⇒ α² + β² = (α + β)² - 2αβ.
⇒ α² + β² = (-8)² - 2(6).
⇒ α² + β² = 64 - 12.
⇒ α² + β² = 52.
Products of the zeroes of the quadratic polynomial.
⇒ αβ = c/a.
⇒ (α²) x (β²) = (αβ)² = (6)² = 36.
As we know that,
Formula of quadratic polynomial.
⇒ x² - (α + β)x + αβ.
Put the values in the equation, we get.
⇒ x² - (52)x + (36).
⇒ x² - 52x + 36.
MORE INFORMATION.
Conjugate roots.
(1) = If D < 0.
One roots = α + iβ.
Other roots = α - iβ.
(2) = If D > 0.
One roots = α + √β.
Other roots = α - √β.
Given :-
If α and β are the zeroes of polynomial x² + 8x + 6
To Find :-
Polynomial whose zeroes are α² and β²
Solution :-
On comparing the given equation with ax² + bx + c. We get
a = 1
b = 8
c = 6
Sum of zeroes = -b/a
Sum = -(8)/1
Sum = -8
Product of zeroes = c/a
Product = 6/1
Product = 6
Now
New Sum of zeroes = α² + β²
α² + β²
(α + β)²
α² + 2αβ + β²
(α + β)² - 2αβ
(-8)² - 2(6)
64 - 12
52
New Product of zeroes = α² × β²
α² × β²
(α × β)²
(6)²
36
Standard form of quadratic polynomial = x² - (α + β)x + αβ
Quadratic polynomial = x² - (52)x + (36)
Quadratic polynomial = x² - 52x + 36