Math, asked by bharath3271, 2 months ago

find a quadratic polynomial the sum and product of whose zeroes are √2 and -3/2 respectively. find the zeroes also​

Answers

Answered by Anonymous
54

Given

  \sf \to \: sum \:  of \: zeroes \:  =  \sqrt{2}

 \sf \to \: product \: of \: zeroes \:  =   \dfrac{ - 3}{2}

To Find

 \sf \to \: quadratic \:  \: polynomial \:  \: and \: zeroes

We have

 \sf \to( \alpha  +  \beta ) =  \sqrt{2}

 \sf \to( \alpha  \beta ) =  \dfrac{ - 3}{2}

General Equation of polynomial

  \to \sf {x}^{2}  - ( \alpha  +  \beta )x +  \alpha  \beta  = 0

Put the value

 \sf \to \:  {x}^{2}  -  \sqrt{2} x -  \dfrac{3}{2}  = 0

Now Simplify the equation

 \sf \to 2 {x}^{2}  -  2\sqrt{2} x - 3 = 0

Now Find the zeroes

 \sf \to \: 2 {x}^{2}  - 3 \sqrt{2} x +  \sqrt{2} x - 3 = 0

 \sf \to \:  \sqrt{2} x( \sqrt{2} x - 3) + 1( \sqrt{2} x - 3) = 0

 \sf \to \: ( \sqrt{2} x + 1)( \sqrt{2} x - 3) = 0

 \sf \to \:  \sqrt{2} x + 1 = 0 \: and \:  \sqrt{2} x - 3 = 0

 \sf\to \: x =  \dfrac{ - 1}{ \sqrt{2} }  \:  \: and \: x =  \dfrac{3}{ \sqrt{2} }

answer

\sf\to \: x =  \dfrac{ - 1}{ \sqrt{2} }  \:  \: and \: x =  \dfrac{3}{ \sqrt{2} }

Answered by Anonymous
32

Answer:

Given :-

Sum of zeroes = √2

Product of zeroes = -3find a quadratic polynomial the sum and product of whose zeroes are √2 and -3/2 respectively. find the zeroes also/2

To Find :-

  • Quardtic polynomial
  • Zeroes

Solution :-

We know that

 \tt \:  {x}^{2}  -( \alpha  +  \beta)x + \alpha \beta

Here,

 \sf \:  \alpha  +  \beta  =  \sqrt{2}

 \sf \:  \alpha  \beta  =  \dfrac{c}{a}

 \sf \:  \alpha  \beta  =  \dfrac{ - 3}{2}

Now,

By putting the value

 \tt \:  {x}^{2}  - ( \sqrt{2} )x +  \dfrac{ - 3}{2}

 \tt \:  {x}^{2}  -  \sqrt{2x }  +  \dfrac{ - 3}{2}

 \sf \: 2 {x}^{2}  -\sqrt{2}  x- 3

Now

Finding zeroes

2x² - 3√2x + √2x - 3 = 0

√2x(√2x - 3) + 1(√2x - 3) = 0

Now,

By taking √2x as common

(√2x + 1) (√2x - 3)

So,

Either

x = -1/√2

or,

x = 3/√2

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