Question

Determine the zeroes of the polynomial


`(\sqrt{ {x }^(2) - 4x + 3} ) + ( \sqrt{ {x}^(2) - 9} ) - ( \sqrt{4 {x }^(2) - 14x + 6 } )`
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Answer 1

3,7/6

Explanation:

`(√(x^2-4x+3) )+(\sqrt{x^(2) -9} )-(√(4x^2-14x+6) )\\=(√(x^2-x-3x+3) )+(√((x^2-3^2))-(√(4x^2-2x-12x+6))\\ =(√(x(x-1)-3(x-1)) )+√((x+3)(x-3))-√(2x(2x-1)-6(2x-1)) \\=√((x-1)(x-3))+√((x+3)(x-3)) -√(2(2x-1)(x-3)) \\=√(x-3) (√(x-1) +√(x+3) -√(2(2x-1)) )\\`

`√(x-3) =0~gives~x=3\\or~√(x-1) +√(x+3) -√(2(2x-1)) =0\\or~ √(x-1) +√(x+3) =√(2(2x-1)) \\squaring\\x-1+x+3+2√(x-1) √(x+3) =2(2x-1)\\2x+2+2√((x-1)(x+3)) =4x-2\\2√(x^2-x+3x-3) =2x-4\\√(x^2+2x-3) =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=(7)/(6)`