Chapter 12: Radicals and More Connections to Geometry By Marcy Schaeffer and Chris Simmons

12.1 Functions Involving Square Roots Key Words square root function: y=√x domain: all the possible input values of a function (all the numbers you plug into the equation) range: the collection of all output values of a function (all the possible answers after you plug a number into an equation)

X (Domain) Input Y (Range) Output Function: y=x²

12.2 Operations with Radical Expressions Key Words simplest form of a radical expression: form when all of the possible square roots are no longer part of the equation The distributive property can be used to simplify the expression when they have the same radicand (i.e. if all parts of the have a √3).

Examples!! Yay. 4√3 - √27= 4√3 - √(9 x 3) 4√3 - √27= 4√3 - √(9 x 3) = 4√3 - √9 x √3 = 4√3 - 3√3 = 4√3 - √9 x √3 = 4√3 - 3√3 = √3 = √3 Factor one of the numbers under the square root only if one of the factors is the same as another number under a square root. Then distribute the square root to the factors. Finally, simplify.

Examples (cont.) √2 x √8= √16= 4 √2 x √8= √16= 4 Multiply the numbers under the square root as if there was no square root sign. Then, put the square root sign over the number and simplify. 3/√5 = 3/√5 x (√5/√5) 3/√5 = 3/√5 x (√5/√5) If an expression has a square root sign on the bottom, multiply the top and the bottom by the number with the square root sign.

12.3 Solving Radical Equations Key Words radical: the square root sign extraneous solutions: solutions that when plugged back into the original equation do not work. These solutions come from squaring both sides. You can make sure you do not have any of these solutions by plugging the answers back into the original equation.

Ejemplos If a=b then a=b. If a=b then a²=b². √x – 7= 0 √x – 7= 0 √x= 7 √x= 7 (√x) = (7) (√x) ²= (7)² x=49 x=49 First simplify the equation by moving everything but the x to one side of the equal sign. Then, square both sides to get rid of the square root sign over the x. By squaring both sides, x is left by itself.

Ejemplos (cont.) Solve x + 20 = x and check for extraneous solutions. ❷ Square each side of the equation. ( x ) ² = x² ❸ Simplify the equation. x + 20 = x² ❹ Write the equation in standard form. X² – x – 20 = 0 ❺ Factor the equation. (x – 5)(x + 4) = 0 ❻ Use the zero-product property to solve for x. x = 5 or x = –4 CHECK Substitute 5 and –4 in the original equation. CHECK X = 5 CHECK X = –4 x + 20 = x x + 20 = x ?= 5 – ?= –4 5 = 5 4 ≠ –4 ANSWER ➧ The only solution is 5, because x = –4 does not satisfy the=original equation.

Solving Radical Equations with Extraneous Solutions First, square both sides of the equation sign to get rid of the square root sign above the x. Then, move everything to one side of the equal side so it equals 0. Solve for x by factoring or by using the quadratic formula. Finally, plug these solutions for x into the original equation and see if the equation works. If one of the solutions you found for x does not work, then that value of x is not a solution for the equation.

12.4 Rational Exponents Key Words cube root of a: If b³ =a, then b is called a cube root of a. rational exponents: a^(m/n)= (a^(1/n))^m

Properties of Rational Exponents a^m x a^n= a^(m+n) a^m x a^n= a^(m+n) If you are multiplying numbers of the same base (i.e. “a”) with exponents, then add the exponents of the numbers. (a^m)^n (a^m)^n If you are multiplying a number with an exponent and just an exponent, then multiply the exponents. (ab)^m (ab)^m If you are multiplying numbers in parentheses by an exponent, then distribute the exponents to the numbers in the parentheses.

Examples 5^(1/3) x 5^(2/3) 5^(1/3) x 5^(2/3) Since the base of the exponents is the same (the 5), then add the exponents. 1/3+2/3=3/3=1. Thus the answer is 5^1 or 5. (7^1/3)^6 (7^1/3)^6 Since you are multiplying a base with an exponent and just an exponent, simply multiply the exponents. 1/3 x 6=2 so the answer is 7 or 49. Since you are multiplying a base with an exponent and just an exponent, simply multiply the exponents. 1/3 x 6=2 so the answer is 7² or 49.

12.5 Completing the Square Key Words: completing the square: To complete the square of x + bx, add the square of half of the coefficient (the number in front of) of x, that is, add (b/2). completing the square: To complete the square of x² + bx, add the square of half of the coefficient (the number in front of) of x, that is, add (b/2)². quadratic formula:

Methods for Solving Quadratic Equations Finding square roots Finding square roots Graphing Graphing Using the Quadratic Formula Using the Quadratic Formula Factoring (if easy to do so) Factoring (if easy to do so) Completing the Square (most useful when number in front of x is 1 and the number in front of the x is even) Completing the Square (most useful when number in front of x² is 1 and the number in front of the x is even)

Solving a Quadratic Equation x + 10x= 24 x² + 10x= 24 x + 10x + 5 = x² + 10x + 5² = ² (x + 5)=49 (x + 5)²=49 x + 5= 7 or x + 5= -7 x + 5= 7 or x + 5= -7 x= 2 or -12 x= 2 or -12 First, make sure all x’s are on one side of the equal side, while all the numbers without an x are on the other. Then, divide the number in front of the x by 2. Square this number and add it to both sides of the equation. Then, find the factors of the quadratic equation and solve it by factoring or by using the quadratic equation.

Remember! When completing the square, always add the number divided by two square to both sides of the equation.

12.6 The Pythagorean Theorem and Its Converse Key Words: theorem: a statement that can be proven true Pythagorean theorem: If a triangle is a right triangle, then the sum of the squares of the lengths of legs a and b equals the square of the length of the hypotenuse c. a + b= c Pythagorean theorem: If a triangle is a right triangle, then the sum of the squares of the lengths of legs a and b equals the square of the length of the hypotenuse c. a² + b²= c² hypotenuse: the side of a triangle across from the right angle, the longest side legs: the other two sides of a triangle that are not the hypotenuse converse: a statement in which the hypothesis and conclusion are interchanged

Using the Pythagorean Theorem Usually the two sides or legs of a right triangle are labeled a or b, and the hypotenuse is labeled c.

Example…..who knew? By using the Pythagorean theorem, just determine the given values of a, b, and c. Remember, a and b are the legs of the triangle and c is the hypotenuse. Then, plug the values into the equation and solve for the unknown value.

A harder example A right triangle has one leg that is 7 inches longer than the other leg. The hypotenuse is 17 inches. Find the unknown lengths. SOLUTION a² + b² = c² Write Pythagorean theorem. x² + (x + 7) ² = 172 Substitute for a, b, and c. x² + x² + 14x + 49 = 289 Simplify. 2x² + 14x – 240 = 0 Write in standard form. X² + 7x – 120 = 0 Divide each side by 2. (x – 8)(x + 15) = 0 Factor. x = 8 or x = –15 Zero-product property ANSWER ➧ Length is positive, so the solution x = –15 is extraneous. The sides have lengths of 8 inches and = 15 inches. CHECK Substitute 8 and 15 for the lengths of the legs in the Pythagorean theorem. a² + b² = c² ?= = 289

Converse of the Pythagorean Theorem If a triangle has side lengths a, b, and c such that a + b = c, then the triangle is a right triangle. If a triangle has side lengths a, b, and c such that a² + b² = c², then the triangle is a right triangle. For example, by plugging side lengths 15, 20, and 25 into the Pythagorean theorem, it shows that the triangle is a right triangle; the numbers on both sides of the equal sign are the same. Remember that the longest side is the hypotenuse, or c.

12.7 The Distance Formula Key Words: distance formula: Given the two points (x1, y1) and (x2, y2), the distance between these points is given by the formula: distance formula: Given the two points (x1, y1) and (x2, y2), the distance between these points is given by the formula:

Distance Formula.

Distance Formula Example Use the distance formula to find the distance between points (37, 122) and (34, 117)

Tough Distance Formula Example A player kicks a soccer ball from a position that is 8 yards from a sideline and 10 yards from a goal line. The ball lands at a position that is 45 yards from the same sideline and 38 yards from the same goal line. How far was the ball kicked? SOLUTION Assuming the kicker is left of the goalie, the ball is kicked from the point (8, 10) and lands at the point (45, 38). Use the distance formula. d = (x2 – x1)² + (y2 – y1) ² Write the formula. = (45 – 8)² + (38 – 10) ² Substitute. = Simplify. = Evaluate powers. = 2153 Add. ≈ 46.4 Find the square root. ANSWER ➧ The ball was kicked about 46.4 yards.

X-pla-nay-shun To find the distance between two points, first determine which point is the “first” point and which point is the “second” point (which point is “first” or “second” does not matter). Then, plug in the x and y values into the distance formula. Simplify and solve.

12.9 The Midpoint Formula Key Words: midpoint: the point on a segment that is the same distance from both the endpoints midpoint formula:

Midpoint Example Given the points ( 1, -2 ) and ( -3, 5 ), find the midpoint of the line segment joining them. Given the points ( 1, -2 ) and ( -3, 5 ), find the midpoint of the line segment joining them. Using the midpoint formula, label the points as x1 = -1, y1 = -2, x2 = -3, and y2 = 5. To find the midpoint M of the line segment joining the points, use the midpoint formula: Using the midpoint formula, label the points as x1 = -1, y1 = -2, x2 = -3, and y2 = 5. To find the midpoint M of the line segment joining the points, use the midpoint formula:

Overview To find the midpoint of a line, determine the endpoints of the segment. Of these endpoints, determine the “first” and “second” points. Plug these values of x and y into the equation.

12.9 Logical Reasoning: Proof Key Words: postulate/axiom: properties that mathematicians accept without proof counterexample: used to show that a general statement is false indirect proof: proof by assuming the statement is false conjecture: a statement that is thought to be true but has not been proven true

The Basic Axioms of Algebra Let a, b, and c be real numbers. Axioms of Addition and Multiplication Closure: a + b is a real number Commutative: a + b = b + a Associative: (a + b) + c= a + (b + c) Identity: a + 0 = a, 0 + a= a Inverse: a + (-a) = 0 Axiom Relating Addition and Multiplication Distributive: a(b + c) = ab + ac Axioms of Equality Addition: If a = b, then a + c = b + c Multiplication: If a = b, then ac = bc Substitution: If a = b, then a can be substituted for b.

The End Get excited! You just learned about equations with the square root sign and with exponents, the Pythagorean theorem, the distance formula, and the midpoint formula. How fun.