Math, asked by adilkhangaming, 10 months ago

Find quedratic polynomial if sum of product of zeros are 2√3and-√3

Answers

Answered by PRANEETHA
0

Step-by-step explanation:

Hope it is useful ........

Attachments:
Answered by BrainlyConqueror0901
4

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\therefore{Quadratic\:eqn\:\:x^{2}-2\sqrt{3}x-\sqrt{3}=0}}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

 \green{\underline \bold{Given:}} \\  \tt: \implies Sum \: of \: zeroes  = 2 \sqrt{3}  \\  \\  \tt:  \implies Product \: of \: zeroes =  -  \sqrt{3}  \\  \\  \red{\underline \bold{To \: Find:}} \\  \tt: \implies Quadratic \: polynomial = ?

• According to given question :

 \tt{Let \: zeroes \: be \:  \alpha  \: and \:  \beta } \\  \\  \tt \circ \:  \alpha  +  \beta  =2 \sqrt{3}  \\  \\  \tt \circ \: \alpha  \times \beta  =  -  \sqrt{3}   \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies  { x }^{2}  - (Sum \: of \: zeores)x + (Product \: of \: zeroes)  = 0\\  \\  \tt:  \implies  {x}^{2}  - ( \alpha  +  \beta )x + ( \alpha  \beta ) = 0 \\  \\  \text{Putting \: given \: value} \\  \tt:  \implies  {x}^{2}  - 2 \sqrt{3} x  + ( -  \sqrt{3} ) = 0 \\  \\  \tt:  \implies  {x}^{2}  - 2 \sqrt{3} x -  \sqrt{3}  = 0 \\  \\  \green{\tt\therefore quadratic \: eqn  =  {x}^{2}  - 2 \sqrt{3} x - \sqrt{3}  = 0} \\  \\   \blue{\bold{Some \: related \: formula}} \\  \orange{ \tt  \circ  \: D =  {b}^{2} -  4ac} \\  \\  \orange{ \tt  \circ  \: x =  \frac{ - b \pm \sqrt{D} }{2a} }

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