Math, asked by skfayaiz96769, 11 months ago

Find the quadratic polynomial,for the zeroes alpha,beeta given in each case.
{¡}.2,-1
{¡¡}.root3,-root3
{¡¡¡}.1/4,-1
{¡v}.1/2,3/2

Answers

Answered by utopian08
0

Answer:

plz use the formula

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

let me q no 2 for you

ans is

 {x}^{2}  - 3

Step-by-step explanation:

 {x}^{2}  - ( \sqrt{3}  + ( -  \sqrt{3} ))x +  \sqrt{3}  \times ( -  \sqrt{3} )

 {x}^{2}  - ( \sqrt{3}  -  \sqrt{3} )x -  \sqrt{3}  \times  \sqrt{3}

Answered by SUBASHRAJ
0

Answer:

plz use the formula

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

2

−(α+β)x+αβ=0

let me q no 2 for you

ans is

{x}^{2} - 3x

2

−3

Step-by-step explanation:

{x}^{2} - ( \sqrt{3} + ( - \sqrt{3} ))x + \sqrt{3} \times ( - \sqrt{3} )x

2

−(

3

+(−

3

))x+

3

×(−

3

)

{x}^{2} - ( \sqrt{3} - \sqrt{3} )x - \sqrt{3} \times \sqrt{3}x

2

−(

3

3

)x−

3

×

3

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