Algebra Know-It-ALL

372 Quadratic Equations with Real Roots

Such a root is said to have multiplicity 2. If the above polynomial expressions are placed on the

left sides of equations with 0 on the right, we get these quadratics:

x^2 + 2 x+ 1 = 0

x^2 − 2 x+ 1 = 0

x^2 + 4 x+ 4 = 0

9 x^2 + 12 x+ 4 = 0

Respectively, they factor into:

(x+ 1)^2 = 0

(x− 1)^2 = 0

(x+ 2)^2 = 0

(3x+ 2)^2 = 0

Because the expressions on the left sides of the above equations are equal to 0, we can take the

square roots of both sides in each case without “plus-or-minus” ambiguity. This gives us the

following first-degree equations, respectively:

x+ 1 = 0

x− 1 = 0

x+ 2 = 0

3 x+ 2 = 0

The solutions to these first-degree equations are easy to find:

x=− 1

x= 1

x=− 2

x=−2/3

These are the roots of the original quadratics. We can write down the solution sets as the

single-element sets {−1}, {1}, {−2}, and {−2/3}.

Positive numbers on the right

Now we’ll depart slightly from the polynomial standard form. Let’s use the same perfect

squares as above, but set the right sides to positive numbers rather than 0. Here are the new

equations:

x^2 + 2 x+ 1 = 1

x^2 − 2 x+ 1 = 4

x^2 + 4 x+ 4 = 16

9 x^2 + 12 x+ 4 = 25