Find the quadratic polynomial whose zeroes are 6 + √ 3 3 and 6 − √ 3 3 . Options: A . &nbsp x 2 − 16 x + 9 B . &nbsp 3 x 2 − 12 x + 11 C . &nbsp 6 x 2 − 4 x + 1 D . &nbsp 7 x 2 − 4 x + 11

Answer: Option B : B Let α & β be the roots. Sum of zeroes, α + β = 6 + √ 3 3 + 6 − √ 3 3 = 12 3 = 4 Product of zeroes, α β = 6 + √ 3 3 × 6 − √ 3 3 = 6 2 − ( √ 3 ) 2 9 α β = 36 − 3 9 = 33 9 = 11 3 Required polynomial f ( x ) = [ ( x 2 − ( α + β ) x + ( α β ) ] = [ ( x 2 − 4 x + 11 3 ] Multiplying by 3 ⇒ f ( x ) = 3 x 2 − 12 x + 11 ∴ the required polynomial is 3 x 2 − 12 x + 11 .

Find the quadratic polynomial whose zeroes are 6+√33 and 6&#x

Question

Find the quadratic polynomial whose zeroes are $\frac{6 + \sqrt{3}}{3} and \frac{6 - \sqrt{3}}{3} .$

Options:

A .

x^{2} - 16 x + 9

B .

3 x^{2} - 12 x + 11

C .

6 x^{2} - 4 x + 1

D .

7 x^{2} - 4 x + 11

Answer: Option B
:
B

Let $α & β$ be the roots.
Sum of zeroes, $α + β = \frac{6 + \sqrt{3}}{3} + \frac{6 - \sqrt{3}}{3}$ $= \frac{12}{3} = 4$
Product of zeroes, $α β = \frac{6 + \sqrt{3}}{3} \times \frac{6 - \sqrt{3}}{3} = \frac{6^{2} - (\sqrt{3})^{2}}{9} α β = \frac{36 - 3}{9} = \frac{33}{9} = \frac{11}{3}$
Required polynomial $f (x) = [(x^{2} - (α + β) x + (α β)] = [(x^{2} - 4 x + \frac{11}{3}]$
Multiplying by 3
$\Rightarrow f (x) = 3 x^{2} - 12 x + 11$
$∴$ the required polynomial is $3 x^{2} - 12 x + 11$ .