Question

find zeroes :of the following quadratic polynomial and verify relationship between the zeroes and the coefficient

Ч√3x^2 +5x-2√3​

Answer 1

Step-by-step explanation:

${4 \sqrt{3}x }^{2} + 5x - 2 \sqrt{3} = 0 \\ 4 \sqrt{3} {x}^{2} + (8 - 3)x - 2 \sqrt{3} = 0 \\ 4 \sqrt{3} {x}^{2} + 8x - 3x - 2 \sqrt{3} = 0 \\ 4x( \sqrt{3} x + 2) - \sqrt{3} ( \sqrt{3} x + 2) = 0 \\ (4x - \sqrt{3} )( \sqrt{3} x + 2) = 0 \\ \\ 4x - \sqrt{3} = 0 \: and \sqrt{3} x + 2 = 0 \\ x = \frac{ \sqrt{3} }{4} andx = \frac{ - 2}{ \sqrt{3} } \\ \alpha = \frac{ \sqrt{3} }{4} and \beta = \frac{ - 2}{ \sqrt{3} } \\ \\ \alpha + \beta = \frac{ - b}{a} \\ \frac{ \sqrt{3} }{4} + ( \frac{ - 2}{ \sqrt{3} } ) = \frac{ - 5}{4 \sqrt{3} } \\ \frac{3 - 8}{4 \sqrt{3} } = \frac{ - 5}{4 \sqrt{3} } \\ \frac{ - 5}{4 \sqrt{3} } = \frac{ - 5}{4 \sqrt{3} } \\ \\ \alpha \beta = \frac{c}{a} \\ \frac{ \sqrt{3} }{4} \times \frac{ - 2}{ \sqrt{3} } = \frac{ - 2 \sqrt{3} }{4 \sqrt{3} } \\ \frac{ - 1}{2} = \frac{ - 1}{2}$

Answer 2

Answer:

4

3

x

2

+5x−2

3

=0

4

3

x

2

+(8−3)x−2

3

=0

4

3

x

2

+8x−3x−2

3

=0

4x(

3

x+2)−

3

(

3

x+2)=0

(4x−

3

)(

3

x+2)=0

4x−

3

=0and

3

x+2=0

x=

4

3

andx=

3

−2

α=

4

3

andβ=

3

−2

α+β=

a

−b

4

3

+(

3

−2

)=

4

3

−5

4

3

3−8

=

4

3

−5

4

3

−5

=

4

3

−5

αβ=

a

c

4

3

×

3

−2

=

4

3

−2

3

2

−1

=

2

−1

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