Question

Form a quadratic. polynomial one whose zero is 2+root 5 and product of zeroes is -18.​

Answer 1

Given

• One Zero of Quadratic equation is 2√5.
• Product of Zeros = (-18).

To Find

• Quadratic Equation

Solution

_______________________

• Let alpha and beta are the Zeros of Polynomial f(x).

$\sf \alpha = 2 \sqrt{5}$

$\sf{Product \: of \: Zeros = - 18}$

We know that

$\\ \sf{ \: \: \: \: \implies \alpha \times \beta = Product \: of \: Zeros }$

$\sf{ \: \: \: \: \implies 2 \sqrt{5} \times \beta = - 18 }$

$\sf{ \: \: \: \: \implies \beta = \dfrac{ - 18}{2 \sqrt{5} } }$

$\sf{ \: \: \: \: \implies \beta = \dfrac{ - 9}{\sqrt{5} } }$

• We know that a Quadratic Equation when the Sum and Product of its Zeros are given by.

$\bf{f(x) = x²-(Sum \: of \: Zeros)x + Product \: of \: Zeros }$

$\sf{f(x) = x²- \left(2 \sqrt{5} + \dfrac{ - 9}{ \sqrt{5} } \right)x + - 18 }$

$\sf{f(x) = x²- \left(\dfrac{ \sqrt{5}( 2 \sqrt{5} )- 9}{ \sqrt{5} } \right)x - 18 }$

$\sf{f(x) = x²- \left(\dfrac{ 2 \times 5 - 9}{ \sqrt{5} } \right)x - 18 }$

$\sf{f(x) = x²- \left(\dfrac{ 10 - 9}{ \sqrt{5} } \right)x - 18 }$

$\sf{f(x) = x²- \dfrac{ x}{ \sqrt{5} } - 18 }$

Hence,

• when One Zero of Quadratic equation is 2√5. Product of Zeros is (-18) then the Quadratic Equation will be x²-x/√5 -18.

Answer 2

$\huge{\textbf{{{\color{navy}{Aղ}}{\purple{Տա}}{\pink{ҽᖇ♤࿐}}{\color{pink}{:}}}}}$

alpha=root5

alpha x beta=-2root5

root5 x alpha=-2root5

beeta=-2

alpha+beeta=root5-2

p(x)=x^2-root5x+2x-2root5

hope it's correct

${\huge{\boxed{\tt{\color {red}{thanks❀✿°᭄}}}}}$