Question

find the zeroes of x²-root 3x-6 I give u 5 star ⭐
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Answer 1

x²-√3x-6

= x²-(2√3-√3)x-6

= x²-2√3x+√3x-6

= x(x-2√3)+√3(x-2√3)

= (x-2√3)(x+√3)

x²-√3x-6 = 0

=> (x-2√3)(x+√3) = 0

=> (x-2√3) = 0 => (x+√3) = 0

=> x = 2√3 => x = -√3

the zeros of ax²+bx+c = {-b+(√b²-4*a*c)}/2a and {-b-(√b²-4*a*c)}/2a

the zeros of x²-√3x-6 =

in this equation b = (-√3)

a = 1

c = (-6)

{-(-√3)+(√(-√3)²-4*1*(-6))}/2*1

= {√3+(√3+24)}/2

= {√3+√27}/2

= {√3+3√3}/2

= (4√3)/2

= 2√3

another zero

{-(-√3)-(√(-√3)²-4*1*(-6))}/2*1

= {√3-(√3+24)}/2

= {√3-√27}/2

= {√3-3√3}/2

= (-2√3)/2

= -√3