Help with Quadratic Equations by completing the square

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 3K views
Maddsnicole
Messages
1
Reaction score
0
I understand how to solve these equations when the square is on this side of the equal sign: x2 + 8x + 7 = 27

But when the square is on the other side, I am thrown. Like this one...
x2 = 14x - 33

The solutions manual shows the next step as the following, but what do you do to get to this point?
x2 - 14x + 49 = -33 + 49
 
Mathematics news on Phys.org
We are given:

$$x^2=14x-33$$

Subtract through by $14x$:

$$x^2-14x=-33$$

Take the coefficient of the linear term which is -14, divide by 2 to get -7, then square to get 49, and so add 49 to both sides:

$$x^2-14x+49=-33+49$$

Does this make sense?
 
Maddsnicole said:
I understand how to solve these equations when the square is on this side of the equal sign: x2 + 8x + 7 = 27

But when the square is on the other side, I am thrown. Like this one...
x2 = 14x - 33

The solutions manual shows the next step as the following, but what do you do to get to this point?
x2 - 14x + 49 = -33 + 49

I would not have followed the solution manuals next step. From the question

x2 = 14x - 33

I would first move everything over to the left hand side (because this is where the x2 is).

When you do this the terms you move from one side to the other need to change signs (positive becomes negative, negative becomes positive) so you should get this:

x2 - 14x + 33 = 0

From here I would complete the square as normal.