How do you solve #x^3 + 147 = 3x^2 + 49x#?

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Answer 1 · 2015-11-27T22:43:56

Subtract #147+49x# from both sides, then identify common factors on each side to find:

#x = 3#, #7# or #-7#

Subtract #147+49x# from both sides to get:

#x^3-49x = 3x^2-147#

That is:

#x(x^2-49) = 3(x^2-49)#

So either #x = 3# or #x^2-49 = 0#, giving #x = +-sqrt(49) = +-7#

#color(white)()#

Alternatively and more systematically:

Subtract the right hand side from the left to get:

#x^3-3x^2-49x+147 = 0#

Let #f(x) = x^3-3x^2-49x+147#

Factor by grouping:

#x^3-3x^2-49x+147#

#=(x^3-3x^2)-(49x-147)#

#=x^2(x-3)-49(x-3)#

#=(x^2-49)(x-3)#

#=(x^2-7^2)(x-3)#

#=(x-7)(x+7)(x-3)#

...using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

So #f(x) = 0# has roots #x = 7#, #x = -7# and #x = 3#