Question

Write the polynomial whose zeros are √3,-√3
with explanation.

correct answer=marked as BRAINLIEST​

Answer 1

To Find :-

• Quadratic polynomial.

Solution :-

Given,

• Zeroes of quadratic polynomial is √3 and -√3.

[ First find out sum of roots ]

$\longrightarrow$ Sum of roots = √3 + (-√3)

$\longrightarrow$ Sum of roots = √3 - √3

$\longrightarrow$ Sum of roots = 0

.°. α + β = 0 $\green\bigstar$

[ Now, find out product of roots ]

$\longrightarrow$ Product of roots = √3 × -√3

$\longrightarrow$ Product of roots = -√3 × 3

$\longrightarrow$ Product of roots = -3

.°. αβ = -3 $\orange\bigstar$

As we know that,

Quadratic polynomial ;

↪ x² - (α + β)x + αβ

[ Put the values ]

$\longrightarrow$ x² - (0)x + (-3)

$\longrightarrow$ x² - 0x - 3

$\longrightarrow$ x² - 3 $\red\bigstar$

Therefore,

The required quadratic polynomial is x² - 3.

Answer 2

★QUESTION:-

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Write the polynomial whose zeros are √3,-√3

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★SOLUTION:-

$\sf \bullet \: \: Let \: \alpha \: and \: \beta \: be \: the \: zeros \: of \: the \: polynomial$

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Now

$\sf \longrightarrow Sum \: of \: the \: zeros = \alpha + \beta\\\sf = \sqrt{3} + ( - \sqrt{3)} \\ \bf = 0 \: \:$

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$\sf \longrightarrow Product \: of \: the \: zeros = \alpha \times \beta \\\sf \:\:\:= \sqrt{3} \times( - \sqrt{3}) \\ \bf \: \: \: \: = -3$

Now, we have the the given polynomial

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$\underline { \boxed{ \bf \longmapsto \: {x}^{2} + ( \alpha + \beta)x + ( \alpha \beta)}}$

Putting the value of sum and product of zeros we get,

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$\implies \sf {x}^{2} + 0x - 3$

$\sf \: or \implies \sf {x}^{2} - 3$

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Therefore , x²-3 is the required polynomial

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Let's verify it :-

We know that in equation ax²+bx+c,where a=1,b=0 and c=3 from the required polynomial (x²-3)

•Sum of the zeros =-b/a=-0/1=0

•Product of the zeros=c/a=3/1=3