Question

C) If one of the root of the quadratic equation is (3-V5) then form the
quadratic equation​

Answer 1

Step-by-step explanation:

Remember that :

Roots like this occurs in pair

If one of the roots is $3 - \sqrt{5}$ then another root will be $3 + \sqrt{5}$

To form quadratic equation...

${x}^{2} - (\alpha + \beta)x + \alpha \times \beta = 0 \\ \\ Where \: \alpha \: and \: \beta \: are \: roots$

$Let \: \\ \\ \alpha = 3 - \sqrt{5} \\ \\ \beta = 3 + \sqrt{5} \\ \\ \implies \alpha + \beta = 3 \cancel{- \sqrt{5}} + 3 \cancel{+ \sqrt{5}} \\ \\ \implies \alpha + \beta = 3 + 3 \\ \\ \boxed{ \implies \alpha + \beta = 6 } \\ \\ \\ \implies \alpha \beta = (3 - \sqrt{5}) \times (3 + \sqrt{5}) \\ \\ We \: know.. \\ \\ \boxed{ (a + b) \times (a - b) = {a}^{2} - {b}^{2}} \\ \\ Using \: the \: formula.... \\ \\ \implies \alpha \beta = (3 - \sqrt{5}) \times (3 + \sqrt{5}) \\ \\ \implies \alpha \beta = {(3)}^{2} - {(\sqrt{5})}^{2} \\ \\ \implies \alpha \beta = 9 + 5 \\ \\ \implies \boxed{\alpha \beta = 14 } \\ \\ \implies {x}^{2} - (\alpha + \beta)x + \alpha \beta = 0 \\ \\ \implies {x}^{2} - (6)x + 14 = 0 \\ \\ \large\blue{\boxed{\bold{ \implies {x}^{2} - 6x + 14 = 0 }}}$