# a) x 2 8x = 25 x 2 8x + 16 = (x 4) 2 = 41 x = 4 ± 41 x + 1 = ± 6 e) x 2 = 5 c) 2x 2 + 2x 7 = 0 2x 2 + 2x = 7 x 2 + x = 7 2

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1 Solving Quadratic Equations By Square Root Method Solving Quadratic Equations By Completing The Square Consider the equation x = a, which we now solve: x = a x a = 0 (x a)(x + a) = 0 x a = 0 x + a = 0 x = a x = a x = ±a Because we solve certain equations the same way all the time, we can now take a shortcut. Square Root Rule: You may take the square root of both sides of an equation provided you use ± on one side. This allows us to solve many different types of equations. Ex 1: x = 5 x = ± 5 WARNING: Ex : (x + ) = 5 x + = ± 5 x = ± 5 If you use the square root rule, you had better remember the ±. Forgetting the ± will not be tolerated!! 1. Put the equation in a form such that the quadratic and linear terms are on one side of the equation and the constant term is on the other side.. Make sure that the quadratic coefficient is one. Divide both sides by the coefficient if necessary. 3. Make the critical calculations: a) Take the linear coefficient. b) Take half of it (or divide it by ). c) Square it. d) Add this result to both sides of the equation. 4. Factor one side (which must be a perfect square trinomial) and simplify the other. 5. Take the square roots of both sides of the equation. (Don't forget that one side needs ±) 6. Solve the resulting equation and simplify. 7. Check your answers. Examples: Solve the following equations by the square root method: Examples: Solve the following equations by completing the square: a) x = 4 x = ± 4 x = ± b) y = 5 y = ± 5 c) z = 18 z = ± 18 z = ±3 d) w = 9 w = ± 9 w = ±3i a) x 8x = 5 x 8x + 16 = (x 4) = 41 x 4 = ± 41 b) 3x + 6x 15 = 0 3x + 6x = 15 x + x = 5 x + x + 1 = x = 4 ± 41 (x + 1) = 6 x + 1 = ± 6 e) x = 5 x = ± 5 x = ±i 5 f) x + 4 = 0 x = 4 x = ± 4 x = ±i g) (x 3) = 5 x 3 = ±5 x = 3 ± 5 x = 8, h) (y + ) = 9 y + = ±3 y = ±3 y = 1, 5 c) x + x 7 = 0 x + x = 7 x = 1 ± 6 d) x + 5x 3 = 0 x + 5x = 3 x + x = 7 x + 5 x = 3 i) (z 4) = 5 z 4 = ± 5 z = 4 ± 5 j) (w + 1) = 4 w + 1 = ±i w = 1 ± i k) (x + 1) = 3 x + 1 = ± 3 x = 1 ± 3 x = 1 ± 3 x + x = (x + 1 ) = 15 4 x + 1 = ± 15 x = 1 ± 15 x + 5 x = (x ) = x = ±7 4 x = 5 4 ± 7 4 x = 3, 1

2 e) x 4x + 7 = 0 x 4x = 7 x 4x = (x 1) = 7 x 1 = ± 7 x = 1 ± 6 f) x 3x + 1 = 0 x 3x = 1 x 3x = (x 3 ) = 5 4 x 3 = ± 5 x = 3 ± 5 g) 3x 6x + 15 = 0 3x 6x = 15 x x = 5 x x + 1 = 5+1 (x 1) = 4 h) x + x + = 0 x + x = x + x + 1 = +1 (x + 1) = 1 x + 1 = ±i i) x x + 13 = 0 x x = 13 x x +1 = 13+1 (x 1) = 1 x 1 = ±i 1 x 1 = ±i x = 1 ± i x = 1 ± i x = 1 ± i 3 Examples: Solve the following using the quadratic formula: a) x 5x + 3 = 0 a = 1, b = 5, c = 3 x = b ± b 4ac a = ( 5) ± ( 5) 4(1)(3) (1) = 5 ± 5 1 = 5 ± 13 5 ± 13 b) 5x x 4 = 0 a = 5, b = 1, c = 4 x = b ± b 4ac a = ( 1) ± ( 1) 4(5)( 4) (5) = 1 ± = 1 ± = 1 ± 9 10 x = 1, 4 5 Solving Quadratic Equations By The Quadratic Formula c) x + x 4 = 0 d) 3 4 x 1 x = 1 3 Briefly stated, the quadratic formula tell us that if ax + bx + c = 0 with a 0, then x = b ± b 4ac a. You should memorize this formula. These are the steps you should use to solve an equation: 1. Put the equation in standard form. This means you must get a zero on one side of the equation.. Identify the corresponding values of a, b, and c. 3. Evaluate the expression b ± b 4ac a with these values of a, b, and c. 4. Simplify the expression. 5. Check your answers. The quantity b 4ac is called the discriminant and gives us the following information about the solution of the equation. If b 4ac > 0 then there are two real solutions. If b 4ac = 0 then there is one real solution. If b 4ac < 0 then there are no real solutions. (there are two complex solutions) a = 1, b =, c = 4 x = b ± b 4ac a = () ± () 4(1)( 4) (1) = ± = ± 0 = ± 5 = ( 1 ± 5) x = 1 ± 5 1( 3 4 x 1 x ) = 1( 1 3 ) 9x 6x = 4 9x 6x 4 = 0 a = 9, b = 6, c = 4 x = b ± b 4ac a = ( 6) ± ( 6) 4(9)( 4) (9) = 6 ± = 6 ± = 6 ± = 6(1 ± 5) 18 1 ± 5 3

3 e) 1 4 x 1 x = 1 4 4( 1 4 x 1 x ) = 4( 1 4 ) x x = 1 x x + 1 = 0 a = 1, b =, c = 1 x = b ± b 4ac a = ( ) ± ( ) 4(1)(1) (1) = ± 4 4 = ± 0 = 1 f) 7x x + 1 = 0 a = 7, b =, c = 1 x = b ± b 4ac a = ( ) ± ( ) 4(7)(1) (7) = ± = ± 4 14 = ± i 6 14 = (1 ± i 6) 14 1 ± i 6 7 Examples: Solve the following equations: a) x 4 7x + 1 = 0 (x 4)(x 3) = 0 b) (x + ) (x + ) 1 = 0 u u 1 = 0 x 4 = 0 x 3 = 0 (u 4)(u + 3) = 0 x = 4 x = 3 u 4 = 0 u + 3 = 0 x = ± x = ± 3 u = 4 u = 3 x + = 4 x + = 3 x =,, 3, 3 x = x = 5 x =, 5 c) x /3 4x 1/3 1 = 0 u 4u 1 = 0 (u 7)(u + 3) = 0 d) x x 1/ = 0 u u = 0 (u )(u + 1) = 0 u 7 = 0 u + 3 = 0 u = 0 u + 1 = 0 u = 7 u = 3 u = u = 1 x 1/3 = 7 x 1/3 = 3 x 1/ = x 1/ = 1 x = 7 3 x = ( 3) 3 x = = 4 x = ( 1) = 1 x = 343 x = 7 x = 4 x = 343, 7 g) x 3x + 4 = 0 a =, b = 3, c = 4 x = b ± b 4ac a = ( 3) ± ( 3) 4()(4) () = 3 ± = 3 ± ± i 3 4 h) 3x + 5 = 0 a = 3, b = 0, c = 5 x = b ± b 4ac a = (0) ± (0) 4(3)(5) (3) = ± 60 6 = ±i 15 6 ±i 15 3 e) x 4 7x + 10 = 0 u 7u + 10 = 0 (u 5)(u ) = 0 u 5 = 0 u = 0 u = 5 u = x = 5 x = 1 x = 5 1 x = x = 1 5 x = ± 1 5 x = ± 5 5 x = 1 x = ± 1 x = ± x = ± 5 5, ± f) x 5 16x = 0 x(x 4 16) = 0 x(x 4)(x + 4) = 0 x = 0 x 4 = 0 x + 4 = 0 x = 4 x = 4 x = ± x = ±i x = 0, ±, ±i g) x x = 48 x x + 48 = 0 (x + 4)(x + 1) = 0 x + 4 = 0 x + 1 = 0 x = 4 x = 1 x = ±i x = ±i 1 x = ±i 3 x = ±i, ±i 3

4 Solving Word Problems ( Super Solvers Use C.A.P.E.S.) 1. Read the problem carefully. (Reread it several times if necessary). Categorize the problem type if possible. (Is it a problem of numerical expression, distance rate time, cost profit, or simple interest type?) 3. Decide what is asked for, and A ssign a variable to the unknown quantity. Label the variable so you know exactly what it represents. 4. Draw a Picture, diagram, or chart whenever possible!! 5. Form an E quation (or inequality) that relates the information provided. 6. Solve the equation (or inequality). 7. Check your solution with the wording of the problem to be sure it makes sense. distance rate time: distance = rate time (d = r t) cost profit: profit = revenue cost (P = R C) simple interest: interest = principal * rate (i = p r) total cost total cost = cost / item * # items work problems 1 / A + 1 / B = 1 / together Example : The sum of the squares of three consecutive positive odd integers is 83. Find the integers. 1 st odd integer: n nd odd integer: n+ 3 rd odd integer: n+4 sum of squares = 83 n + (n+) + (n+4) = 83 n + n + 4n n + 8n + 16 = 83 3n + 1n 63 = 0 3(n + 4n 1) = 0 3(n 3)(n + 7) = 0 n = 3, 7 7 not positive n = 3 only solution integers: 3, 5, & 7 consecutive integers n, n+1, n+,... or n, n+, n+4,... Pythagorean Theorem a + b = c Example 1: The product of two consecutive integers is 0. Find the integers. 1 st integer: n nd integer: n+1 product = 0 n(n + 1) = 0 n + n = 0 n + n 0 = 0 (n 4)(n + 5) = 0 n = 4, 5 Example 3: The length of a rectangle is meters more than its width. If the diagonal is 10 meters, what are the dimensions of the rectangle? 10 w + w Pythagorean theorem w + (w+) = 10 w + w + 4w + 4 = 100 w + 4w 96 = 0 (w + w 48) = 0 (w + 8)(w 6) = 0 n = 4 then n+1 = 5 n = 5 then n + 1 = 4 4 & 5 or 5 & 4 w = 8, 6 8 make no sense length: 8 meters width: 6 meters

5 Example 4: In two hours, a motorboat can travel 8 miles down a river and return 4 miles back. If the river flows at a rate of miles per hour, how fast can the boat travel in still water? Example 6: A grocer mixes \$9.00 worth of Grade A coffee with \$1.00 worth of Grade B coffee to obtain 30 pounds of a blend. If Grade A costs 30 a pound more than Grade B, how many pounds of each were used? down up d = r t 8 x+ 4 x 8/(x+) 4/(x ) charge = price quantity Grade A 9 9/x x Grade B 1 1/(30 x) 30 x 8 x x 8 (x+)(x ) x total time = hours x = = (x+)(x ) 8(x ) + 4(x + ) = (x 4) 8x x + 8 = x 8 0 = x 1x = x(x 6) x = 0, 6 6 mph 9 x 9 10x(30 x) x = 1 30 x = 10x(30 x) 90(30 x) = 10x + 3x(30 x) x = 10x + 90x 3x 3x 300x = 0 3(x 100x + 900) = 0 3(x 10)( x 90) = 0 x = 10, 90 only x = 10 makes sense 10 pounds of Grade A 0 pounds of Grade B 1 30 x Example 5: Two employees together can prepare a large order in hr. Working alone, one employee takes three hours longer than the other. How long does it take each person working alone? one x+3 hours other x hours together hours Solving Polynomial or Rational Inequalities Never multiply both sides of an inequality by an expression involving the variable. To do so involves a complicated procedure you are not ready to handle yet. WARNING: Multiplying both sides of an inequality by an expression involving the variable will not be tolerated!! To solve polynomial or rational inequalities: 1 x x x(x + 3) 1 x x = 1 = x(x + 3) 1 x + (x + 3) = x(x + 3) x + x + 6 = x + 3x 0 = x x 6 0 = (x 3)(x + ) x = 3, only x = 3 makes sense 6 hours for one 3 hours for other Step 1: Put the inequality into Standard Form. (Zero on one side) Step : Simplify the other side of the inequality. Step 3: Factor the resulting expression. Step 4: Identify all critical values. (They cause the numerator or denominator to be zero. Step 5: Create a sign chart with values that cause the numerator to be zero marked with 0 and values which cause the denominator to be zero marked with (*). This divides the real line into intervals. Step 6: Check each interval using test points to see if the rational expression is positive or negative. If positive, mark with + ; if negative, mark with. Step 7: Choose the appropriate interval remembering never to choose any value marked with (*).

6 Examples: Solve the following inequalities and use Mathematics Writing Style: a) x x > 8 b) x x h) 6 x + 1 i) 5 x x x x 8 = 0 (x 4)(x + ) > 0 x 13x (x 3)(x 5) 0 6 x x x / 5 6 (x + ) x + 5 (x 7)(x + 3) x x + 3 x < or x > 4 3/ x 5 6 x x x 4x 1 x + 3 c) x 3 + 4x 5x < 0 x(x + 4x 5) < 0 d) x 3 4x x 3 4x 4 x x + x 4x + 4 x + 3 x(x + 5)(x 1) < x < 5 or 0 < x < 1 x(x 4) x(x )(x + ) x 0 or x (*) < x 4 (x ) x + 3 (*) x > 3 e) x + 3 x 4 f) x 4 x Graphing Quadratic Functions + 0 (*) x 3 or x > 4 (x )(x + ) x (*) x or 1 < x g) x + 8 x 5 + < 0 x (x 5) x 5 x x 10 x 5 < 0 < 0 3x x 5 < (*) + /3 5 /3 < x < 5 y = x y = x + 1 y = x + 4 y = x 3 typical graph typical graph moved up 1 unit typical graph moved up 4 units typical graph moved down 3 units

7 y = x typical graph y = (x 1) y = (x 4) y = (x + 3) typical graph moved right 1 unit typical graph moved right 4 units typical graph moved left 3 units y = x y = x y = 5x typical graph typical graph inverted typical graph much thinner, inverted y = 0.x typical graph much wider, inverted Graphing Quadratic Functions A quadratic function is a function whose rule is a quadratic polynomial. That is, it can be written in the form or f(x) = ax + bx + c, a 0. f(x) = a(x h) + k, a 0. The graph of such a quadratic function is a parabola with the following information: y = x y = x y = 5x y = 0.5x y = 0.x typical graph typical graph slightly thinner typical graph much thinner typical graph slightly wider typical graph much wider up if a > 0 1) The parabola opens down if a < 0 thinner if a > 1 ) The parabola is wider if 0 < a < 1 3) The vertex of the parabola has coordinates (h, k) where h = b a k = f(h) = c ah 4) The axis of symmetry of the parabola is the line x = h Remember: If you have the quadratic function given in standard form, you can always rewrite it into the vertex form using the above formulas for h and k!

8 Note that it is much easier to graph when the function has the form f(x) = a(x h) + k. Here are some suggestions for graphing quadratic functions. Step 1: Write in standard form and identify a, b, and c. Step : Calculate h: h = b a Step 3: Calculate k: k = f(h) OR k = c ah Step 4: Find all of the intercepts. Step 5: Plot sufficiently many points and make use of the intercepts and the vertex (h, k) Step 6: Draw the smooth graph. Example: Graph f(x) = x 4x 3. y Vertex: (, 7) Intercepts: (0, 3), (+ 7,0), ( 7,0) Example: Graph f(x) = x x + 3. Find the vertex and any intercepts. y

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