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

Hey!

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Refer pic for the zeroes and the coefficients are -:

a = 4 √3
b = 5
c = - 2√3

We know, [ Note - @ = alpha , ß = beta ]

So, @ + ß = - b/a

= - 5 / 4√3

Sum of zeroes = - b/a

-2/3 + √3/4 = - 5 / 4√3


@ × ß = c/a

= - 2 √3 / 4 √3

= - 2/4

= - 1/2

Product of zeroes = c/a

-2/3 + √3/4 = -1/2



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Hope it helps...!!!

Answer 2

Hey there

• Given

$4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3}$

Here a = 4 √3 , b = 5 , c = - 2√3

where , a × c = ( 4√3 ) × ( -2 √3 ) = 24

$= 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} = 0\\ \\ = 4 \sqrt{3} {x}^{2} + 8x - 3x - 2 \sqrt{3} = 0 \\ \\ =(4 \sqrt{3} {x}^{2} + 8x) - (3x + 2 \sqrt{3} ) = 0\\ \\ = 4x( \sqrt{3}x + 2) - \sqrt{3} ( \sqrt{3} x + 2) = 0\\ \\ = (4x - \sqrt{3} )( \sqrt{3x} + 2) = 0 \\ \\ = x = \frac{ \sqrt{3} }{4} \: or \: \frac{ -{2} }{ \sqrt{3} }$
now ,

$\frac{ - 2}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{ - 2 \sqrt{3} }{3}$

Hope this helps you ☺