Find the roots of x^2-4x-8 by the method of completing square
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Answered by
3
x²-4x-8=0
x²-2*4/2x+(4/2)²-(4/2)²-8=0
x²-2×2x+2²-2²-8=0
(x-2)²-4-8=0
(x-2)²=4+8
(x-2)²=12
x-2=+-4root3
x=2+-4root3
x=2+4root3
x=2-4root3
x²-2*4/2x+(4/2)²-(4/2)²-8=0
x²-2×2x+2²-2²-8=0
(x-2)²-4-8=0
(x-2)²=4+8
(x-2)²=12
x-2=+-4root3
x=2+-4root3
x=2+4root3
x=2-4root3
janaksinghrajpoot:
sorry
it's not 4root3 it's 2root3
Answered by
2
x = (4+√48)/2 and x = (4-√48)/2
i.e. x = 2+2√3 and x= 2-2√3
i.e. x = 2+2√3 and x= 2-2√3
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