Question

find the zeroes of the polynomial root 3x square+10x+7 rpot 3 and verify the relationship between zeroes ans their coefficients

Answer 1

p(x) = √3x2+10x+7√3

= √3x+3x+7x+7√3

= (√3x+7)(x+√3)

= √3x+7 = 0 and x+√3 = 0

= x = -7/√3 and x = -√3

verification by α and β,

let α = -7/√3 and β = -√3

sum of zeros = α+β = -7/√3+(-√3)

= -7/√3-√3

= -7-3/√3

= -10/√3

product of zeroes = αβ

= -7/√3 . -√3

= 7.

verification by coefficients,

sum of zeros = -b/a

= -10/√3.

product of zeros ,

= c/a

= 7√3/√3

= 7.

hence relationship is verified.

Answer 2

Hello dear...

Solution here ..✌✌
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Let the given Polynomial be denoted by f(x) . Than ,

$f(x) = \sqrt{3} {x}^{2} + 10x + 7 \sqrt{3} \\ = > \sqrt{3} {x}^{2} + 3x + 7x + 7 \sqrt{3} \\ = > \sqrt{3}x(x + \sqrt{3} ) + 7(x + \sqrt{3} ) \\ = > (x + \sqrt{3} )( \sqrt{3} x + 7) \\ \\ f(x) = 0 \\ \\ = > (x + \sqrt{3} ) = 0 \: \: or \: \: ( \sqrt{3} x + 7) = 0 \\ = > x = - \sqrt{3} \: \: \: \: \: \: \: \: and \: \: x = \frac{ - 7}{ \sqrt{3} }$
so the zeros of f (x) are -√3 and -7/√3

We have to verify
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$sum \: \: of \: zeros \: = ( - \sqrt{3} + \frac{ - 7}{ \sqrt{3} } ) = \frac{ - 10}{ \sqrt{3} } = \frac{coefficient \: \: of \: x)}{coefficiet \: \: of \: {x}^{2} } \\ \\ product \: \: of \: zeros = \frac{ - 7}{ \sqrt{3} } \times - \sqrt{3} = 7 = \frac{constant \: \: term}{coefficient \: \: of \: {x}^{2} }$
Hence verified ..

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Hope it's helps you.
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