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