Question

find the qudratic equation whose root 3and 1/3​

Answer 1

sum of zeroes 3+1/3 =10/3

product of zeroes 3×1/3=1

quadratic equation =x2 -sumx+product

therefore equation is x square - 10/3x+ 1

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Answer 2

✡Required Answer:-

☞Given:-

• $\sf \footnotesize{ \alpha \: = 3}$
• $\sf \footnotesize{ \beta \: = \frac{1}{3} }$
• $\sf \footnotesize{Here \: \alpha \: and \: \beta \: are \: the \: zeroes. \: }$

☞To find:-

• The quadratic polynomial having 3 and 1/3 as zeroes

☞Solution:-

Now,

$\sf \footnotesize{Sum \: of \: zeroes = \alpha + \beta = 3 + \frac{1}{3} = \frac{10}{3} }$

And,

$\sf \footnotesize{Product \: of \: zeroes = \alpha \times \beta = 3 \times \frac{1}{3} = 1}$

Now, we know that

$\sf \footnotesize{Quadratic \: equation = {x}^{2} + (sum \: of \: zeroes)x + product \: of \: zeroes}$

$\sf \footnotesize{ \therefore Required \: quadratic \: equation = {x}^{2} + ( \alpha + \beta )x + \alpha \beta }$

$\sf \footnotesize{ \implies {x}^{2} + \frac{10}{3} x + 1 = 0}$

$\sf \footnotesize{ \implies \: \: \frac{ 3 {x}^{2} + 10x + 3}{3} = 0}$

$\sf \footnotesize{ \implies \: 3 {x}^{2} + 10x + 3 = 0}$

Hence, 3x^2 + 10x + 3 = 0 is the required quadratic equation.✔