In algebra, a quadratic equation is any equation having the form where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic.

Quadratic equations are usually called second degree equations, which mean that the second degree is the highest degree of the variable that can be found in the quadratic equation. 

Solving Quadratic equations appear on most College standardized tests and some High School Proficiency exams

If we are unsure how to rewrite quadratic equations so we can solve it using factoring method, we can use the formula for quadratic equation:

We write x_1,2 because it represents 2 solutions of the equation. Here is a practice question :

We see that a is 3, b is -10 and c is 3.

We use these numbers in the equation and do some calculations.

Notice that we have + and -, so x₁ is for + and x₂ is for -, and that’s how we get 2 solutions.

1. Find 2 numbers that sum to 21 and the sum of the squares is 261.

a. 14 and 7
b. 15 and 6
c. 16 and 5
d. 17 and 4

2. Using the factoring method, solve the quadratic
equation: x² + 4x + 4 = 0
a. 0 and 1
b. 1 and 2
c. 2
d. -2

3. Using the quadratic formula, solve the quadratic
equation: x – 31/x = 0
a. -?13 and ?13
b. -?31 and ?31
c. -?31 and 2?31
d. -?3 and ?3

4. Using the factoring method, solve the quadratic
equation: 2x² – 3x = 0
a. 0 and 1.5
b. 1.5 and 2
c. 2 and 2.5
d. 0 and 2

5. Using the quadratic formula, solve the quadratic
equation: x² – 9x + 14 = 0
a. 2 and 7
b. -2 and 7
c. -7 and –2
d. -7 and 2

Answer Key

1. B
The numbers are 15 and 6.
x + 7 = 21 => x = 21 -7
x² + y² = 261
(21 – 7 )² + y² = 261
441 – 42y + y² + y² = 261
2y² – 42y + 180 = 0

y² – 21y + 90 = 0
y_1,2 = 21 + ?441 – 360/2
y_1,2 = 21 + ?81/2
y_1,2 = 21 + 9/2
y₁ = 15
y2 = 6
x₁ = 21 = y₁ = 21 – 15 = 6
x₂ = 21 – y₂ = 21 – 6 = 15

2. D

-2

3. B

4. A

0 and 1.5
2x² – 3x = 0
x(2x – 3)
x = 0 or 2x – 3 =0
x = 0 or x = 3/2
x = 0 or x = 1.5

5. A
2 and 7