Question

Find a quadratic polynomial whose zeroes are 2√5-3√2 and 2√5+3√2.

Answer 1

Answer:

${x}^{2} - 4 \sqrt{5} x + 2 = 0$

Step-by-step explanation:

$s = 2 \sqrt{5} - 3 \sqrt{2} + 2 \sqrt{5} + 3 \sqrt{2} = 4 \sqrt{5} \\ \\ p = (2 \sqrt{5} - 3 \sqrt{2} )(2 \sqrt{5} + 3 \sqrt{2} ) \\ \\ = {(2 \sqrt{5} )}^{2} - {(3 \sqrt{2}) }^{2} \\ \\ = 20 - 18 \\ \\ p= 2 \\ \\ required \: equation \: is \: \: \\ \\ {x}^{2} - sx + p = 0 \\ {x }^{2} - 4 \sqrt{5} x + 2 = 0$

Answer 2

Answer:

x²-4√5x+2 = 0

Step-by-step explanation:

quadratic equation,

x²-(α+β)x+αβ = 0

α = 2√5-3√2 β = 2√5-3√2

α+β = 2√5-3√2+2√5+3√2

= 4√5

αβ = (2√5-3√2)(2√5+3√2)

= (2√5)²-(3√2)²

= 20-18 = 2

x²-(α+β)x+αβ = 0

x²-4√5x+2 = 0