Question 1091387
{{{(1+8x)^(0.5)=1+4x-8x^2+32x^3}}}
So then substituting,
{{{(1+8(0.01))^(0.5)=1+4(0.01)-8(0.01)^2+32(0.01)^3}}}
{{{sqrt(1.08)=1+(0.04)-(0.0008)+(0.000032)}}}
{{{sqrt(1.08)=1.039232)}}}
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{{{1+8x=3}}}
{{{8x=2}}}
{{{x=1/4}}}
However, the Taylor series expansion at {{{x=0}}} only converges when {{{abs(8x)<1}}}
{{{abs(x)<1/8}}}