Algebra Know-It-ALL

The left sides can be factored exactly as before, getting

(x+ 1)^2 = 1

(x− 1)^2 = 4

(x+ 2)^2 = 16

(3x+ 2)^2 = 25

If we take the square roots of both sides in each case here, remembering to include both the
positive and negative results, we obtain

x+ 1 =±(1)1/2

x− 1 =±(4)1/2

x+ 2 =±(16)1/2

3 x+ 2 =±(25)1/2

These simplify to the following “double-barreled” first-degree equations:

x+ 1 =± 1

x− 1 =± 2

x+ 2 =± 4

3 x+ 2 =± 5

Let’s think of these as equation pairs:

x+ 1 = 1 or x+ 1 =− 1

x− 1 = 2 or x− 1 =− 2

x+ 2 = 4 or x+ 2 =− 4

3 x+ 2 = 5 or 3x+ 2 =− 5

When we solve these four pairs of first-degree equations, we get the following results, which
are the roots of the original quadratics:

x= 0 or x=− 2

x= 3 or x=− 1

x= 2 or x=− 6

x= 1 or x=−7/3

We can write the solution sets respectively as {0,−2}, {3,−1}, {2,−6}, and {1,−7/3}. I’ll let you
plug these values into the original quadratics to be sure we’ve found the right roots! Here are
the original quadratics once again, in order:

x^2 + 2 x+ 1 = 1

x^2 − 2 x+ 1 = 4

Completing the Square 373