If a and ? are the zeroes of the quadratic polynomial f(x) = x ^ 2 + x - 2 then find a polynomial whose zeroes are 2alpha + 1 and 2beta + 1
Solution:
Given: α and β are zeroes of the quadratic polynomial f(x) = x2 + x − 2.
To find zeroes: f(x) = 0
x2 + x − 2 = 0
⇒ x2 + 2x − x − 2 = 0
⇒ x(x + 2) − 1(x + 2) = 0
⇒ (x + 2)(x − 1) = 0
⇒ x = 1 or −2
∴ α = 1 and β = −2
⇒ 2α + 1 = 3 and 2β + 1 = −3
Thus, the equation of polynomial with zeros 3 and −3 is given by:
P(x) = (x + 3)(x − 3) = x2 − 9 [∵ (a + b)(a − b) = a2 − b2]
Given: α and β are zeroes of the quadratic polynomial f(x) = x2 + x − 2.
To find zeroes: f(x) = 0
x2 + x − 2 = 0
⇒ x2 + 2x − x − 2 = 0
⇒ x(x + 2) − 1(x + 2) = 0
⇒ (x + 2)(x − 1) = 0
⇒ x = 1 or −2
∴ α = 1 and β = −2
⇒ 2α + 1 = 3 and 2β + 1 = −3
Thus, the equation of polynomial with zeros 3 and −3 is given by:
P(x) = (x + 3)(x − 3) = x2 − 9 [∵ (a + b)(a − b) = a2 − b2]