Asked by Anonymous.
Use quadratic formula
(3x/x-1)+(2/x)=4
the solution set is ( , )
(3x/x-1)+(2/x)=4
the solution set is ( , )
Answers
Answered by
Steve
(3x/x-1)+(2/x)=4
3x*x + 2(x-1) = 4x(x-1)
3x^2 + 2x-2 = 4x^2-4x
x^2-6x+2 = 0
x = 3±√7
3x*x + 2(x-1) = 4x(x-1)
3x^2 + 2x-2 = 4x^2-4x
x^2-6x+2 = 0
x = 3±√7
Answered by
Anonymous.
Sg. root (14-5x) = x
Answered by
Steve
Surely you can do this one:
√(14-5x) = x
14-5x = x^2
x^2+5x-14 = 0
(x+7)(x-2) = 0
But watch out for extraneous roots. Check both values in the original equation. Hint
√4 = 2
√4 ≠ -2
√(14-5x) = x
14-5x = x^2
x^2+5x-14 = 0
(x+7)(x-2) = 0
But watch out for extraneous roots. Check both values in the original equation. Hint
√4 = 2
√4 ≠ -2
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