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Question
`sqrt3x^2+10x-8sqrt3=0`
Solution
Given:
`sqrt3x^2+10x-8sqrt3=0`
On comparing it with `ax^2+bx+x=0` we get;
`a=sqrt3,b=10 and c=-8sqrt3`
Discriminant D is given by:
`D=(b^2-4ac)`
=`(10)^2-4xxsqrt3xx(-8sqrt3)`
=`100+96`
=`196>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt(D))/(2a)=(-10+sqrt(196))/(2sqrt(3))=(-10+14)/(2sqrt(3))=4/(2sqrt3)=2/sqrt(3)=2/sqrt(3)xxsqrt3/sqrt3=(2sqrt(3))/3`
β=`(-b+sqrt(D))/(2a)=(-10+sqrt(196))/(2sqrt(3))=(-10+14)/(2sqrt(3))=(-24)/(2sqrt(3))=(-12)/(2sqrt(3))(-12)/sqrt(3)xxsqrt3/sqrt3=(-12sqrt(3))/3=4sqrt(3)`
Thus, the roots of the equation are `(2sqrt3)/3` and `-4sqrt3`
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