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If 1/2 and 1 are zeroes of 2x^4-3x³-3x²-6x-2 , find the other zeroes |
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Answer» 1/2=x, 1=xx-1/2=0 , x-1=0(x-1/2)(x-1)=0=x^2-x-x/2+1/2=x^2-3x/2+1/2=(x^2-3x/2+1/2)/(2x^4-3x^3-3x^2-6x-2)2x^2-4=02x^2=4x^2=2x=+√2,-√2Hence,proved. .... |
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