Question

form quadratic equation whose roots are 3 minus under root 7 and 3 + under root 7​

Answer 1

Step-by-step explanation:

Roots are

$3 - \sqrt{7} \: and \: 3 + \sqrt{7}$

so sum of zeros will be....

$= (3 - \sqrt{7} ) + (3 + \sqrt{7} ) \\ = 3 - \sqrt{7} + 3 + \sqrt{7} \\ = 6$

And the product of zeros will be ...

$= (3 - \sqrt{7} )(3 + \sqrt{7} ) \\ = {3}^{2} - { \sqrt{(7)} }^{2} \\ = 9 - 7 \\ = 2$

Now , std. form of quad . eq is

$k( {x}^{2} - (sum \: of \: zeros)x + product \: of \: zeros)$

$= k( {x}^{2} - 6x + 2)$

$= {x}^{2} - 6x + 2$

so, the required quad eq is

${x }^{2} - 6x + 2$

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Answer 2

Answer:

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Step-by-step explanation:

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