If 𝛼 and 𝛽 are zeroes of a polynomial 6 𝑥2

Q) If 𝛼 and β are zeroes of a polynomial 6 x² – 5 x + 1, then form a quadratic polynomial whose zeroes are 𝛼² and 𝛽² .

Ans:

Step 1: Given polynomial equation 6 x² – 5 x + 1 = 0

Comparing with standard polynomial, ax² + b x + c = 0, we get,

a = 6, b = – 5, c = 1

Since, its given that the roots of the polynomial be α and β.

and we know that sum of roots (α + β) = $\frac{- b}{a}$

∴ α + β = $\frac{- (- 5)}{6} = \frac{5}{6}$ …………… (i)

Also, we know that the product of the roots (α x β) = $\frac{c}{a}$

∴ α . β = $\frac{1}{6}$ …………. (ii)

Step 2: The zeroes for new polynomial given as (𝛼² , 𝛽² ):

∵ Sum of the zeroes of new polynomial = (α²+ β² ) = (α + β)² – 2 α . β

By transferring values from equations (i) and (ii), we get:

∴ Sum of the zeroes of new polynomial = $(\frac{5}{6})^2 - 2 \times (\frac{1}{6}) = \frac{13}{36}$

Next, Product of the zeroes of new polynomial = (α²).(β²) = (α . β)²

By transferring values from equations (ii), we get:

∴ Product of the zeroes of new polynomial = $(\frac{1}{6})^2 = \frac{1}{36}$

Step 3: New quadratic polynomial f(x) = x²– (sum of the zeroes) x + (product of the zeroes)

∴ $f(\times) = \times ^2 - \frac{13}{36} \times + \frac{1}{36}$

at f(x) = 0, polynomial is ∴ $\times ^2 - \frac{13}{36} \times + \frac{1}{36} = 0$

∴ 36 X² – 13 X + 1 = 0

Hence, the required quadratic polynomial is 36 X²– 13 X + 1 = 0.

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