How do you solve #2(xsqrt x) + x=8#?

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Answer 1 · 2015-07-18T09:45:10

Can reformulate as a cubic equation:

#4x^3-x^2+16x-64 = 0#

which has an irrational root findable using Cardano's method or similar.

Subtract #x# from both sides of the equation to get:

#2xsqrt(x) = 8-x#

Square both sides to get:

#4x^3 = (8-x)^2 = 64 - 16x+x^2#

Note that squaring can introduce spurious solutions, so we need to check later.

Subtract the right hand side from the left to get:

#f(x) = 4x^3-x^2+16x-64 = 0#

By the rational roots theorem, the only possible rational roots of this cubic are:

#+-64#, #+-32#, #+-16#, #+-8#, #+-4#, #+-2#, #+-1#, #+-1/2#, #+-1/4#

None of these are roots (though #2# comes quite close).

graph{4x^3-x^2+16x-64 [-0.698, 4.302, -1.18, 1.32]}

It is possible to solve #f(x) = 0# algebraically, but probably beyond the scope of your course.

First let #x_1 = x - 1/12#

Then:

#4x_1^3 = 4(x-1/12)^3 = 4(x^3-x^2/4+x/48-1/1728)#

#=4x^3-x^2+x/16-1/432#

So #f(x) = 4x^3-x^2+16x-64#

#= 4x_1^3 + (16-1/16)x+(1/432-64)#

#= 4x_1^3 + 255/16x_1+(1/432-64+255/(16*12))#

This is of the form #ax_1^3+bx_1+c#

Then you can use Cardano's method to solve, finding a solution of the form:

#x_1 = root(3)(A+sqrt(B)) + root(3)(A-sqrt(B))#

hence

#x = root(3)(A+sqrt(B)) + root(3)(A-sqrt(B)) + 1/12#

If you are really interested see: