Question

Find the quadratic polynomial whose zeros are are 3-√3/5 and 3+√3/5​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

We are provided that , 3 - √3/5 and 3 + √3/5 are the zeros of the quadratic polynomial. We need find the quadratic polynomial.

$\dag\:\underline{\tt Let \: \alpha =\dfrac{3 - \sqrt{3}}{5} \:and \: \beta = \dfrac{3 + \sqrt{3}}{5} }$

$\qquad \quad\ddag \: \underline{\textbf{Sum of zeros : }}$

$:\implies\sf Sum \: of \: zeros = \alpha + \beta$

$:\implies\sf Sum \: of \: zeros =\dfrac{3 - \sqrt{3}}{5} +\dfrac{3 + \sqrt{3}}{5}$

$:\implies\sf Sum \: of \: zeros =\dfrac{3 + 3}{5}$

$:\implies \underline{ \boxed{\sf Sum \: of \: zeros = \dfrac{6}{5} }}$

Hence,the sum of zeros is 6.

Now, let's find the product of the zeros :

$\qquad \quad\ddag \: \underline{\textbf{Product of zeros : }}$

$:\implies\sf Product \: of \: zeros = \alpha \beta$

$:\implies\sf Product \: of \: zeros = \Bigg(\dfrac{3 - \sqrt{3}}{5} \Bigg) \times \Bigg( \dfrac{3 + \sqrt{3}}{5} \Bigg)$

$:\implies\sf Product \: of \: zeros = \dfrac{ {(3)}^{2} - (\sqrt{3 })^{2} }{ {(5)}^{2} }$

$:\implies\sf Product \: of \: zeros = \dfrac{ 9 - 3 }{ 25 }$

$:\implies \underline{ \boxed{\sf Product \: of \: zeros = \dfrac{6}{ 25 }}}$

Hence,the product of zeros is 6/25.

$\qquad \quad\ddag \: \underline{\textbf{Quadratic Polynomial : }}$

$\dashrightarrow\:\:\sf p(x) = k\bigg[ x^2 - (\alpha + \beta)x + \alpha \beta\bigg]$

$\dashrightarrow\:\:\sf p(x) = k\bigg[ x^2 - \left( \dfrac{6}{5} \right)x + \dfrac{6}{25} \bigg]$

$\dashrightarrow\:\:\sf p(x) = x^2 - \dfrac{6}{5} x + \dfrac{6}{25}$

$\dashrightarrow\:\:\sf p(x) = 25 \times x^2 - 6+ 5 x\times 6$

$\dashrightarrow\:\:\sf p(x) = 25x^2 - 6+ 30 x$

$\dashrightarrow\:\:\sf p(x) = 25 x^2 - 6 + 30x$

$\dashrightarrow\:\: \underline{ \boxed{\sf p(x) = 25x^2 - 30x+ 6}}$

Answer 2

${\boxed{\underline{ \pink{\tt{Required \: \: answer:-}}}}}$

★GIVEN:-

• $\tt{\alpha = \frac{3 - \sqrt{3} }{5} }$

• $\tt\beta = { \frac{3 + \sqrt{3} }{5} }$

★TO FIND :-

• quadratic polynomial

★SOLUTION:-

Sum of zeros:-

$: \implies \tt{ \alpha \: + \beta = \frac{3 - \sqrt{3} }{5} + \frac{3 + \sqrt{3} }{5} }$

$: \implies \tt{ \alpha \: + \beta = \frac{3 - \sqrt{3} + 3 + \sqrt{3} }{5} }$

$: \implies \tt{ \alpha \: + \beta = \frac{6}{5} }$

NOW,

Product of zeros

$: \implies \tt{ \alpha \beta = ( \frac{3 - \sqrt{3} }{5} })( \frac{3 + \sqrt{3} }{5} )$

$: \implies \tt{ \alpha \beta = \frac{9 - 3}{25} }$

$: \implies \tt{ \alpha \beta = \frac{6}{25} }$

Standard equation is:-

${ \boxed{ \underline{ \mathsf{ \purple{ax^2 + bx + c }}}}}$

$: \longrightarrow {x}^{2} - ( \frac{ - b}{a} )x + \frac{c}{a}$

$: \longrightarrow \bf{ \: \alpha \: + \beta \: = \frac{ - b}{a} = \frac{6}{5} }$

$: \longrightarrow \bf{ \alpha \beta = \frac{c}{a} = \frac{6}{25} }$

Putting the values:-

$: \longrightarrow \sf{ {x}^{2} - \frac{6}{5} x + \frac{6}{25} }$

‎

$\leadsto \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \underline{ \blue{ \sf{ 25 {x}^{2} - 30x + 6}}}}}$

This is the required polynomial.