Question

if
$\alpha \: and \: \beta$
are the roots of the quadratic equation
${x}^{2} + \sqrt{2} x + 3 = 0$
form a quadratic polynomial with zeroes
$1 \div \alpha \: 1 \div \beta$

Answer 1

Step-by-step explanation:

these are the following steps to the answer.

Hope you understand

Answer 2

$Answer \: \\ \\ Given \: Quadratic \: Equation \: \: Is \: \\ \\ x {}^{2} + \sqrt{2} x + 3 = 0 \\ \\ \alpha + \beta = \frac{ - \sqrt{2} }{1} \: \: \: \: and \: \: \alpha \beta = \frac{3}{1} \\ \\ \alpha + \beta = - \sqrt{2} \: \: \: \: and \: \: \: \: \: \alpha \beta = 3 \\ \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ - \sqrt{2} }{3} \\ \\ \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{3} \\ \\ let \: the \: required \: Quadratic \: polynomial \: be \: p(x) \\ \\ p(x) = x {}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x + \frac{1}{ \alpha \beta } \\ \\ p(x) = x {}^{2} - ( \frac{ - \sqrt{2} }{3} )x + 1 \\ \\ p(x) =(3 x {}^{2} + \sqrt{2} x + 1) \frac{1}{3}$