Question

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and the coefficients √3x^2 + 10x +7√3

Answer 1

[tex] \sqrt{3} x^{2} +10x+7 \sqrt{3} =0 \\ \sqrt{3} x^{2}+3x+7x+7 \sqrt{3} =0 \\ \sqrt{3}x(x+ \sqrt{3}) + 7(x+ \sqrt{3} )=0 \\ (x+ \sqrt{3} )( \sqrt{3} x+7)=0 \\ x+ \sqrt{3} =0 \\ \alpha = - \sqrt{3} \\ \sqrt{3} x +7=0 \\ \beta = \frac{-7}{ \sqrt{3} } = \frac{-7}{3} \sqrt{3} [/tex]
relationship between zeroes and coefficients
sum of the roots =$\alpha + \beta$ = $\frac{-(coefficient of (x))}{coefficient of (x^{2}) }$=-b/a
there fore [tex] \alpha + \beta =- \sqrt{3} - \frac{7}{3} \sqrt{3} = \frac{-10}{3} \sqrt{3} \\ \alpha + \beta = -b/a = -10/ \sqrt{3} = \frac{-10}{3} \sqrt{3} [/tex]
product of the roots = $\alpha + \beta$ = constant/ coefficient of $x^{2}$ = c/a
$\alpha \beta = (- \sqrt{3} )( \frac{-7}{ \sqrt{3} } ) \\ \alpha \beta =7$
= c/a = $\frac{7 \sqrt{3} }{ \sqrt{3} } = 7$

hence proved 

hope this helps u 
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Answer 2

Answer:

i dont know

Step-by-step explanation: