Answers

2016-04-24T14:00:50+05:30
 \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}
relationship between zeroes and coefficients
sum of the roots = \alpha + \beta  \frac{-(coefficient of (x))}{coefficient of  (x^{2}) } =-b/a
there fore  \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}
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 
pls mark it as brainliest........



1 5 1
will not mark it brainliest.....i already gave u 50 points for nothing
( ・ิω・ิ) (>ω<)
Get lost
seriously??
Yep