How do you solve #x^2 +4x=21#?

1 Answer
anor277
Mar 26, 2016

By simple factorization. Failing that, by use of the quadratic formula.

Explanation:

We are given that #x^2 + 4x =21#, or #x^2 + 4x - 21 =0#. This is a quadratic equation, which is in principle soluble. Factorization presents an easier method.

#(x+7)(x-3)# #=# #x^2 + 4x - 21# upon expansion.

So, if #(x+7)(x-3)# #=# #0#, then clearly #x# has roots of #-7# OR #+3#. Substitution of either of these values for #x# gives #0#. Other posters may give a more systematic method for solution.

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