Question

form the quadratic polynomial whose zeros are -√2 and √2​

Answer 1

Given that,

Zeros of the quadratic polynomial = $\sqrt{-2}\:and\:\sqrt{2}$

To find,

A quadratic equation

Assumption,

Let the zeros be $\alpha$ and $\beta$ as $\sqrt{-2}$ and $\sqrt{2}$ respectively.

As we know, to form a quadratic equation there's a general form of it in terms of its zeros.

The general form to form a quadratic equation is,

$\boxed{\large{\bold{k(x^2+ (\alpha + \beta)x - (\alpha \times \beta)}}}$

Where k is a constant.

By usin' this, we can get our quadratic equation.

$\boxed{\large{\bold{k(x^2 + (\alpha + \beta)x - (\alpha \times \beta)}}}$

$k[x^2+(\sqrt{-2}x + \sqrt{2}) - (\sqrt{-2} \times \sqrt{2})]$

$k[x^2+(2\sqrt{2})x - (-2)]$

Putting k as 1, then our equation will be,

$\boxed{\bold{\pink{x^2 + 2 \sqrt{2}x + 2}}}$

Answer 2

Step-by-step explanation:

Zeros of the quadratic polynomial = \sqrt{-2}\:and\:\sqrt{2}

−2

and

2

To find,

A quadratic equation

Assumption,

Let the zeros be \alphaα and \betaβ as \sqrt{-2}

−2

and \sqrt{2}

2

respectively.

As we know, to form a quadratic equation there's a general form of it in terms of its zeros.

The general form to form a quadratic equation is,

\boxed{\large{\bold{k(x^2+ (\alpha + \beta)x - (\alpha \times \beta)}}}

k(x

2

+(α+β)x−(α×β)

Where k is a constant.

By usin' this, we can get our quadratic equation.

\boxed{\large{\bold{k(x^2 + (\alpha + \beta)x - (\alpha \times \beta)}}}

k(x

2

+(α+β)x−(α×β)

k[x^2+(\sqrt{-2}x + \sqrt{2}) - (\sqrt{-2} \times \sqrt{2})]k[x

2

+(

−2

x+

2

)−(

−2

×

2

)]

k[x^2+(2\sqrt{2})x - (-2)]k[x

2

+(2

2

)x−(−2)]

Putting k as 1, then our equation will be,

\boxed{\bold{\pink{x^2 + 2 \sqrt{2}x + 2}}}

x

2

+2

2

x+2