Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes andthe coefficents of 4√3x^2+5x-2√3

Answer 1

Heya ✌

Here's your answer friend,

4√3x² + 5x -2√3 = 0

==> 4√3x² + 8x - 3x -2√3 = 0

==> 4x(√3x + 2) -√3 (√3x + 2) =0

==> ( 4x - √3)(√3x + 2) = 0

==> x = √3 / 4 or x = -2 / √3

are the zeroes of above polynomial.

On comparing above equation

we get,

a = 4√3x² , b = 5, c = -2√3

Let α = √3 / 4 and β = -2 / √3

Sum of the zeroes

==> α + β = -b / a

==> √3/4 - 2/√3 = - 5 / 4√3

==> -5 / 4√3 = -5/ 4√3

LHS = RHS

And now,

Product of zeroes,

αβ = c / a

==> √3/4 X -2 /√3 = -2√3/4√3

==> -2√3 / 4√3 = -1/2

==> -1/2 = -1/2

LHS = RHS

Hence, verified

Hope it helps you : )

Answer 2

Hello Dear.

Your Question's answer is given in the above attachment
………………………………………………………………

Here, we have to verify This
$sum \: of \: the \: \: zeroes \: \: = ( \frac{ - 2}{3} + \frac{ \sqrt{3} }{4} ) \\ = \frac{ - 5}{4 \sqrt{3} } = \frac{ - ( coefficient \: \: of \: x)}{(coefficient \: \: of \: {x}^{2} )} \\ \\ product \: \: of \: the \: zeroes \: \: = ( \frac{ - 2}{3} \times \frac{ \sqrt{3} }{4} ) \\ = \frac{ - 1}{2} = \frac{constant \: \: term}{coefficient \: \: of \: {x}^{2} }$
HANCE VERIFIED ✓
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@isharoy688
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