solve this equation x^4+4x=48
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Let f(x)=x⁴+4x-48, then f'(x)=4x³+4.

x=x-f(x)/f'(x) where x on the RHS is a guessed value of x producing (hopefully) a more accurate value of x on the LHS.

Observation shows that there are two zeroes for f(x). One is between -3 and -2 and the other between 2 and 3, because f(-3)=21 and f(-2)=-40 (change of sign); and f(2)=-24 and f(3)=45 (another change of sign). The changes of sign indicate a zero between the x values, that is, there is a value of x such that f(x)=0, where -3<x<-2 and 2<x<3.

Therefore, we can use -2.5 and 2.5 as starting values for x. We call these x₀ and treat them separately.







So consecutive iterations converge to x=-2.772546504.

Let A=-2.772546504.





Another zero is x=2.483876601.

Let B=2.483876601.

With these two zeroes we can create a quadratic:


This quadratic is a factor of f(x) so by dividing by it we arrive at another quadratic which may be able to be solved using the formula for solving quadratics.

Let C=-(A+B) and D=AB. Now we use long division to find the second quadratic:


x²+Cx+D ) x⁴                    +4x      -48


                       -Cx³-Dx²+   4x



We need take the division no further because we now have the required quadratic: x²-Cx+C²-D.

The zeroes are x=(C±√(C²-4C²+4D))/2.

The discriminant 4D-3C²<0 when we calculate C and D from the zeroes A and B, so these remaining zeroes are complex.

Therefore the only real zeroes are x=A=-2.772546504 and x=B=2.483876601.

by Top Rated User (816k points)

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