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

Answer 1

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

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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Answer 2

To solve the equation (x^2 + 4x = 21), you can follow these steps:

  1. Rewrite the equation in standard quadratic form: (x^2 + 4x - 21 = 0).
  2. Factor the quadratic expression: ((x + 7)(x - 3) = 0).
  3. Set each factor equal to zero and solve for (x): (x + 7 = 0 \rightarrow x = -7) and (x - 3 = 0 \rightarrow x = 3).

Therefore, the solutions to the equation are (x = -7) and (x = 3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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