1. If $$x+1=x^{2}$$ and $$x>0$$, then $$2x^{4}$$ is
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By: anil on 05 May 2019 02.29 am
We know that $$x^2 - x - 1=0$$
Therefore $$x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$$
Therefore, $$2x^4 = 6x+4$$ We know that $$x>0$$ therefore, we can calculate the value of $$x$$ to be $$frac{1+sqrt{5}}{2}$$
Hence, $$2x^4 = 6x+4 = 3+3sqrt{5}+4 = 3sqrt{5}+7$$
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Therefore $$x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$$
Therefore, $$2x^4 = 6x+4$$ We know that $$x>0$$ therefore, we can calculate the value of $$x$$ to be $$frac{1+sqrt{5}}{2}$$
Hence, $$2x^4 = 6x+4 = 3+3sqrt{5}+4 = 3sqrt{5}+7$$