Find the quadratic polynomial, sum of whose zeroes is (\(\frac{5}2\))

1.	Find the quadratic polynomial, sum of whose zeroes is (\(\frac{5}2\)) and their product is 1. Hence, find the zeroes of the polynomial.
Answer» Let α and β be the zeroes of the required polynomial f(x). Then (α + β) = \(\frac{5}2\) and αβ = 1 ∴ f(x) = x² - (α + β) x + αβ ⇒ f(x) = x² - \(\frac{5}2\) x + 1 ⇒ f(x) = 2x² – 5x + 2 Hence, the required polynomial is f(x) = 2x² – 5x + 2 ∴ f(x) = 0 ⇒ 2x² – 5x + 2 = 0 ⇒ 2x² – (4x + x) + 2 = 0 ⇒ 2x² – 4x – x + 2 = 0 ⇒ 2x (x – 2) – 1 (x – 2) = 0 ⇒ (2x – 1) (x – 2) = 0 ⇒ (2x – 1) = 0 or (x – 2) = 0 ⇒ x = \(\frac{1}2\) or x = 2 So, the zeros of f(x) are \(\frac{1}2\) and 2.

Find the quadratic polynomial, sum of whose zeroes is (\(\frac{5}2\)) and their product is 1. Hence, find the zeroes of the polynomial.

Answer»

Let α and β be the zeroes of the required polynomial f(x).

Then (α + β) = \(\frac{5}2\) and αβ = 1

∴ f(x) = x² - (α + β) x + αβ

⇒ f(x) = x² - \(\frac{5}2\) x + 1

⇒ f(x) = 2x² – 5x + 2

Hence, the required polynomial is f(x) = 2x² – 5x + 2

∴ f(x) = 0 ⇒ 2x² – 5x + 2 = 0

⇒ 2x² – (4x + x) + 2 = 0

⇒ 2x² – 4x – x + 2 = 0

⇒ 2x (x – 2) – 1 (x – 2) = 0

⇒ (2x – 1) (x – 2) = 0

⇒ (2x – 1) = 0 or (x – 2) = 0

⇒ x = \(\frac{1}2\) or x = 2

So, the zeros of f(x) are \(\frac{1}2\) and 2.