1. Section 9.1 “Properties of Radicals”

2. Section 9.1 “Properties of Radicals”
A radical expression is in simplest form if the following conditions are true:
The rules also apply for simplifying cube roots and nth roots.
No perfect square factors other than
1 are in the radicand.
-No fractions are in the radicand.
No radicals appear in the
denominator of a fraction.

3. Product Property of Radicals
The square (or cube) root of a product equals the product of the square ( or cube) roots of the factors.

4. Quotient Property of Radicals
The square root of a quotient equals the quotient of the square roots of the numerator and denominator.

5. Rationalizing the Denominator
Whenever there is a radical (that is not a perfect square) in the denominator, the radical must be eliminated by rationalizing the denominator.
Need to
rationalize the
denominator
Multiply by 1
Product property
of radicals
Simplify

6. Adding and Subtract Radicals
You can add and subtract radicals that have the same radicands.
Think of as combining ‘like terms’
Look for common radicands
Simplify

7. Multiplying Radical Expressions
You can multiply radical expressions the same way you multiplied monomials and binomials using the distributive property and FOIL.
simplify & combine like terms

8. Conjugates
and are called conjugates which can be used to simplify radical expressions containing the sum or difference involving square roots in a denominator.
Need to use a conjugate to
rationalize the denominator
Multiply by 1
Product property of radicals
Simplify

9. Section 9.3 “Solving Quadratic Equations Using Square Roots”

10. Section 9.3 “Solving Quadratic Equations Using Square Roots”
To use square roots to solve a quadratic equation of the form ax² + bx + c = 0, first isolate x² on one side of the equation to obtain x² = d.
Solving x² = d by Taking Square Roots
Examples
- If d > 0, then x² = d has two solutions:
- If d = 0, then x² = d has one solution:
- If d < 0, then x² = d has no solution:
no solution
no solution

11. Try It Out…
No solution

12. Solve the quadratic equation
Solve the quadratic equation. Round your answer to the nearest hundredth.

13. Section 9.4 “Solving Quadratic Equations by Completing the Square”

14. Section 9.4 “Solving Quadratic Equations by Completing the Square”
For an expression of the form x² + bx, you can add a constant c to the expression so that x² + bx + c is a perfect square trinomial. This process is called
COMPLETING THE SQUARE.
x² + bx
(1) Find one-half of b, the coefficient of x.
(2) Square the result from Step 1.
(3) Add the result from Step 2 to x2 + bx. 2
( ) 2 b 2
( ) 2
x² + bx + b 2

15. Complete the Square. Then factor the trinomial.
x² + 8x
(1) Find one-half of b, the coefficient of x.
(2) Square the result from Step 1.
(3) Add the result from Step 2 to x2 + bx. 2 2 8
( ) 2
( ) 2
x² + 8x + 8 2
x² + 8x + (4)2
Factored trinomial
(x + 4)(x + 4)
x² + 8x + 16
(x + 4)2

16. Solving Quadratic Equations by Completing the Square
The method of completing the square can be used to solve any quadratic equation, once the equation is in the form x² + bx = d.
Complete the Square
Add 64 to both sides of the equation
Write the left side of the equation as a binomial (factor).
Simplify right side of equation.
Take the square root of both sides.
Solve for x.

17. Section 9.5 “Solve Quadratic Equations by the Quadratic Formula”

18. Section 9.5 “Solve Quadratic Equations by the Quadratic Formula”
To solve any quadratic equation of the form ax² + bx + c = 0, you can use the QUADRATIC FORMULA.

19. Interpreting the Discriminant
b2 – 4ac > 0
b2 – 4ac = 0
b2 – 4ac < 0

20. Section 9.6 “Solving Systems of Nonlinear Equations”

21. “Solve Nonlinear Systems by Graphing”
y = 2x2+5x–1
Equation 1
y = x – 3
Equation 2
The lines appear to intersect at the point (-1, -4).
The solution is (-1, -4).

22. “Solve Nonlinear Systems by Substitution”
y = -2x + 3
Equation 1
y = x2 + x – 1
Equation 2
(-2x + 3) = x2 + x – 1
y = x2 + x – 1
Substitute
-2x + 3= x2 + x – 1
0 = x2 + 3x – 4
Factor
0 = (x + 4)(x – 1)
x = -4 and 1
y = -2x + 3
Equation 1
y = -2x + 3
Equation 1
Substitute both values for x into the original equation
y = -2(-4) + 3
y = -2(1) + 3
y = 11
y = 1
The solutions are the points (-4,11) and (1,1).
Substitute both points into the equations to check.

23. “Solve Nonlinear Systems by Elimination!”
Eliminated
Eliminated
y = x2 + x
Equation 1 + -
-y = x – 5
y = x + 5
Equation 2
0 = x2 – 5
5 = x2
±2.24 = x2
y = x2 + x
Equation 1
y = x2 + x
Equation 1
Substitute both values for x into the original equation
y = (2.24)
y = (-2.24)
y ≈ 7.24
y ≈ 2.76
The solutions are the points (2.24,7.24) and (-2.24, 2.76).
Substitute both points into the equations to check.