Quadratic Equations

Question 1

How do you recognize a quadratic equation problem?

Answer 1

They will have a variable raised to an exponent and the same variable to the power of 1.
(X^2)+ 4x -12 = 0
or
(X^2)+ 4x = -12

Question 2

After using FOIL what is the step to solving?

x+4)(x+2

Answer 2

Set each = to 0 and solve for each. These will have 2 answers.

Question 3

If equation has 3 variable expressions and one is set to the power of 3, how do you solve?
x^3+x^2-X = 0

Answer 3

Factor out the x, then solve as a normal quadratic equation.

This question will have 3 solutions.

Question 4

If you recognize perfect squares in the quadratic problem, then solve with absolute value. Remember, there will be 2 answers in a quadratic.

Answer 4

EX.

(x+4)^2 = 64
| x+4 | = 8
| x+4 | = - 8
Solve for both.

Question 5

If a quadratic equation is a perfect square, how many solutions will it have?

Answer 5

1

(x+4)(x+4) = 0

Question 6

Special Quadratic Products

1. (x+y)(x-y) =

Answer 6

1. (x^2) - (y^2)

Question 7

Special Quadratic Products

2. (x+y)^2 =

Answer 7

x^2 + 2xy + y^2

Question 8

Special Quadratic Products

3. (x-y)^2 =

Answer 8

x^2 - 2xy + y^2

Question 9

How to Simplify?
2 + 2 = X…
( a + b ) ( a - b)

Answer 9

Multiply both sides of the equation by a common denominator. (a+b)(a-b).

You end up with 4a = a^2 - b^2 (special product)

Question 10

If the 2 answers to a quadratic can not be determined, consider taking the square root from the very beginning

Answer 10

EX.

(p-3)^2 - 5
When you multiply, you end up with
p^2 -6p +4 = 0

You can’t get this, therefore, just square root the first expression and the 5 after moving it over.

ANS
3 + or - the 5^(1/2)

Question 11

Special Quadratic sneaky values like…

Answer 11

1 - n^4
a - b
81 - 4z^2

Notice, they’re all “minus” and only 2 terms (let this be your hint that this could be a hidden special quad)

(1+n)(1-n)(1+n^2)
(square root a) +/- (square root b)