Question

Find the zeroes of the polynomial

p( x ) = x² - 3 and verify the relationship between the zeroes


and coefficients.

Answer 1

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Answer 2

$\huge\bigstar\underline\mathfrak\red{Answer}$

The zeroes of this polynomial p(x) are √3 and -√3.

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$\huge\bigstar\underline\mathfrak\red{Explanation}$

Given : p(x) = x²-3

To find : The zeroes of this polynomial and verify the relationship between zeroes and coefficients.

Solution : Let p(x) = 0

=> x² - 3 = 0

=> x² = 3

=> x = ±√3

Therefore, the zeroes of this polynomial p(x) are √3 and -√3.

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$\huge\bigstar\underline\mathfrak\red{Verification}$

Let $alpha$ and $beta$ be the zeroes of polynomial p(x),

Also, √3 and -√3 are the zeroes.

So, let $alpha$ be √3 and $beta$ be -√3

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Now, $alpha$ + $beta$ = √3 + ( - √3 )

$alpha$ + $beta$ = 0

$alpha$ × $beta$ = √3 × (-√3)

$alpha$ × $beta$ = -3

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We know that,

Sum of zeroes i.e $alpha$ + $beta$ = -b/a

$alpha$ + $beta$ = -0/1 => 0

Also,

Product of zeroes i.e $alpha$ × $beta$ = c/a

$alpha$ × $beta$ = -3/1 => -3

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Hence verified!