Question

find the quadratic polynomial whose zeros are underoot 2 and 2 underoot 2​

Answer 1

Step-by-step explanation:

$roots \: are \: \sqrt{2} \: and \: 2 \sqrt{2} \\ x = \sqrt{2} \\ = > x - \sqrt{2} = 0 \\ and \: x = 2 \sqrt{2} \\ = > x - 2 \sqrt{2} = 0 \\ \\ (x - \sqrt{2} )(x - 2 \sqrt{2} ) = 0 \\ = > x(x - 2 \sqrt{2} ) - \sqrt{2} (x - 2 \sqrt{2} ) = 0 \\ = > {x}^{2} - 2 \sqrt{2} x - \sqrt{2} x + (2 \sqrt{2 } \times \sqrt{2} ) = 0 \\ = > {x}^{2} - 2 \sqrt{2} x - \sqrt{2} x + (2 \times 2)= 0 \\ = > {x}^{2} - 2 \sqrt{2} x - \sqrt{2} x + 4 = 0 \\ = > {x}^{2} - 3 \sqrt{2} x + 4 = 0$

Answer 2

Answer:

Step-by-step explanation:

Let the zeros are  α and β

so α= √2 and β=2√2

then α+β=√2+2√2

αβ=√2*2√2=4

So the polynomial with α and β as zeros:

f(x)=x²-(α+β)x=αβ

=x²-(√2+2√2)x+4

=x²-6/√2*x+4