# Quadratic Equations Worksheet

1. Factor into at least two factors:

**Answer**

2. Find all real solutions of the following equations:

**Answer** The reduced quadratic equation is t^2+2t+1 = 0 with t = x^3.
Factoring

we obtain t^2+2t+1 = (t+1)^2 = 0 and whence the set of solutions to the reduced

quadratic equation is {−1}. Solving t = x^3 = −1 we obtain {−1} as the set of

solutions to the original equation. [Note that since t = −1 is a root of degree
2

of the reduced quadratic , x = −1 is a root of degree 2 of the original
equation.]

Another way to solve this problem is to observe that x^6+2x^3+1 = (x^3+1)^2 = 0

and therefore the solutions satisfy x^3 = −1, which produces the solution set

{−1}.

**Answer** The reduced quadratic is t^2 + t − 3 = 0 with t = (x − 1)^2. By
the

quadratic formula we have

The solution set of the reduced quadratic is therefore

and to obtain the solution set to the original equation we solve

Observe that since
the above calculation produces two complex

solutions. The solution set of the original equation is therefore

Answer The reduced quadratic equation is 8t^2 + 2t + 2 = 0 with
.

Computing the discriminant shows that since it is 4 − 4(8)(2) < 0 that the

reduced quadratic has no real solutions and therefore the original equation

does not have any real solutions. The solution set is therefore the empty set,
.

**Answer** The reduced quadratic equation is

By the quadratic formula we obtain

The solution set of the reduced quadratic equation is therefore

We need to solve t = x^{−3} for each element of this solution set in order to
obtain

the solutions to the original equation.

The solution set of the original equation is therefore

Answer The reduced quadratic equation is t^2 − 2t − 8 = 0 with t = 2^{x}, which

factors as (t − 4)(t + 2) = 0. This tells us that the original equation factors
as

(2^{x} − 4)(2^{x} + 2) = 0 and either 2^{x} = 4 -> x = 2 or 2^{x} = −2, which is impossible.

The solution set of the original equation is therefore {2}.

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