1.1 Solving a Linear Equation ax + b = 0

Transcription

1 1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b = 0 : (i) move b to the other side (subtract b from both sides) (ii) divide both sides by a Example: Solve x = 0 (i) x- = 0 x = (ii) x x Sometimes certain operations on an equation have to be performed before the steps above can be applied. Those operations are: - Removing parenthesis by using distributive property - Clearing fractions by multiplying both sides of the equation by the LCD of all fractions - Grouping like terms 1 1 Example: Solve ( x 5) 4 (x 1) (i) The LCD of fractions ½ and 1/ is 6. Multiply each term by 6 and simplify (ii) Remove the parenthesis by using the distributive property x = 4x (iii) Combine the like terms (variable on the left, constants on the right) - x = 7 (iv) Divide both sides by the coefficient of x x = Solving quadratic equation ax + bx + c = 0 There are three ways to solve a quadratic equation. (a) Solving by factoring (i) write the equation in the standard from: ax + bx + c = 0 (ii) factor, if possible, the right hand side of the equation (iii) Use the Zero-Product Property: If a b = 0, then a = 0 or b = 0; equate each factor to zero and solve corresponding equation. Example: Solve x(x+4) = 1 (i) Write in the standard form (remove parenthesis and bring all terms to the left) x(x+4) = 1 x + 4x 1 =0 (ii) Factor the right hand side (x+6)(x-) = 0 (iii) Use the Zero-Product Property

2 x + 6 = 0 or x = 0 x = - 6 or x = Solutions: -6,. The solution set is {-6, } (b) Solving by the square root method This method can be applied to equations of the form ax + c = 0 (i) Solve the equation for x ax = -c x = -c/a (ii) The solutions are (by taking the square root of both sides) x c a If c/a is negative, there is no real solution If c/a is positive, there are two solutions Example: Solve x + 6 = 0 x + 6 = 0 x = -6 x = - x 6 no (real) solution (can t have a negative number inside the square root) Example: Solve x 15 = 0 x 15 = 0 x = 15 x = 5 x 5 Solutions: 5, - 5 Example: Solve (x+ 1) = This equation can be solved using the square root method, since by substituting x+1 by u, this equation can be rewritten as u =. (x+1) = x+1 = So, we have x+ 1 = or x + 1 = x 1 x x x 1 1 Solutions: x, x (c) Solving using the quadratic formula ( must be memorized) (i) write the equation in the standard form: ax + bx + c = 0 (ii) Identify a, b, c (iii)use the Quadratic Formula: b x b 4ac a

3 If b 4ac 0, there are two different real solutions If b 4ac 0, there is one real solution If b 4ac 0, there are no real solutions The expression b 4ac is called the discriminant of the equation. It determines the number of real solutions. Remark: Any quadratic equation can be solved using quadratic formula. Example: Solve 4x - 1 = - x (i) Write the equation in the standard form 4x - 1 = - x 4x + x - 1 = 0 (ii) Identify a, b, c a = 4, b = 1, c = - 1 (iii) Use the formula 1 x 1 44( 1) Solutions are x, x Solving radical equations and equations leading to quadratic Radical equation: An equation in which a variable appears inside a radical. To solve a radical equation: (i) Isolate a radical (ii) Raise both sides to a power equal to the index of the radical (iii) Simplify and solve the resulting equation (iv) Check Remark: Checking the apparent solutions of a radical equation is a necessary step as raising both sides of an equation might produce numbers that are not the solutions of the original equation Example: Solve x x 4 x 0 (i) Isolate the radical ( Leave the radical on the left and move all other terms to the right) x x 4 x (ii) square both sides (since index = ) x x 4 x x x 4 x (iii) Simplify and solve the resulting equation x x 4 = x + 4x + 4-5x = x = - /5 (iv) Check

4 ? ? 0 yes 5 5 Solution: -/5? 0 5 Solving equations quadratic in form An equation quadratic in form is an equation that becomes quadratic after using an appropriate substitution Example: Consider (1- y) + 5(1-y) + = 0 Notice that if we replace (1 y) by u (we let 1- y = u) we get the equation u + 5u + = 0. Example: Solve z 1/ 4z 1/4 + 4 = 0 Since ½ = ( 1 4 ) we can re-write the equations as (z 1/4 ) 4z 1/4 + 4 = 0 Now substituting z 1/4 by u (u = z 1/4 ) gives u 4u + 4 = 0 This is a quadratic equation. There is only one solution u =. Going back to the z variable we get z 1/4 4 = or z After raising both sides to the fourth power, we get z = 4 = 16 Check: (16) 1/ 4 (16) 1/4 + 4? 0 4 4() + 4 = 0, yes Solution: z = 16 Solving Polynomial Equations: polynomial = 0, where the polynomial has the degree greater than (i) Write the polynomial in the standard form: polynomial = 0 (ii) Factor the polynomial (iii) Use the Zero-Product Property Example: Solve x + x = x (i) x + x = x x + x x = 0 (ii) x(x + x ) = 0 x(x+)(x-1) = 0 (iii) x = 0 or x + = 0 or x 1 = 0 x = - x = 1 Solutions: 0, -, 1

5 1.5 & 5.4 Solving polynomial and rational inequalities Intervals An interval is a subset of the set of real numbers that can be visualized as a segment (finite or infinite) on the number line If a < b, then - The closed interval [a,b] consists of all real numbers x such that a < x < b. We can graph this set on the number line as follows (the endpoints of the segment are included) - The open interval (a, b) consists of all real numbers x such that a < x < b. Below is the graph of this set (the endpoints are not included) - The half open interval (a, b] is the set of all x for which a < x < b - The half open interval [a, b) is the set of all x such that a < x < b - The infinite interval (a, ) is the set of all x such that x > a - The infinite interval [a, ) is the set of all x such that x > a - The infinite interval (-, a) is the set of all x such that x< a - The infinite interval (-, a] is the set of all x such that x < a - The infinite interval (-, ) is the set of all real numbers Solving a Linear Inequality: ax + b > (>, <, < ) 0 (i) (ii) Move b to the right hand side Divide both sides by a, remembering that when dividing or multiplying an inequality by a negative number, we MUST reverse the inequality

6 Example: Solve x > 0 and write the solution in the interval notation (i) - x > 0 -x > - (ii) Divide both sides by (-). The inequality will be reversed. x x Solution: (, /) Solving a Polynomial Inequality: polynomial > (>, <, <) 0 (i) Write the inequality in the standard form (0 on the right hand side) (ii) Solve the equation : polynomial = 0 (iii) Plot the solution on the number line (iv) The solutions divide the number line into a finite number of intervals. Choose a number in each interval and evaluate the value of the polynomial at each number. (v) If the value of the polynomial at the chosen number is positive (> 0), then the polynomial is positive on the whole interval If the value of the polynomial is negative (< 0), then the polynomial is negative on the whole interval (vi) Choose, as the solution, the intervals on which the polynomial has a desired sign. Use interval notation. Include the endpoints only when the original inequality is < or >. If there are two separate intervals on which the polynomial has a desired sign, use the union sign,, between the intervals. Example: Solve x > x (i) x > x x x > 0 (ii) x x = 0 x(x - 1)= 0 x = 0 or x 1 =0 x = 1 x = 1 1 (iii) (iv) x x -x - (-) (-) = - + = -6 negative (-0.5) - (-0.5)= =.75 positive 0.5 (.5) -(.5)= -.75 negative - = 6 positive (v)

7 (vi) Since the inequality is x x 0, we choose the intervals on which the polynomial is positive and include the endpoints. There are two intervals, so we use the union symbol. Solution: [-1,0] [1,) P Solving a Rational Inequality: (,, ) 0, P, Q are polynomials Q P (i) Write the inequality in the standard form (,, ) 0 Q (ii) Solve the equations: P = 0 and Q = 0 (iii) Plot the solutions on the number line. Place open circle at each solution of Q = 0; those numbers cannot ever be included in the solution set (they make the denominator zero!) (iv) The solutions divide the number line into a finite number of intervals. Choose a number P in each interval and evaluate the value of the expression at each number. Q (v) If the value of Q P is positive (> 0), then Q P is positive on the whole interval If the value of Q P is negative (< 0), then Q P is negative on the whole interval (vi) Choose, as the solution, the intervals on which Q P has a desired sign. Use interval notation. Include the endpoints only when the original inequality is < or >. Remember to never include the endpoint with an open circle! If there are two or more such intervals, use the union sign,. x Example: Solve x 4 (i) x x 4 x 0 x 4 x ( x 4) 0 x 4 x 4 x x 0 x 4 x 10 0 x 4 (ii) Numerator = 0 denominator = 0 x+10 = 0 x- 4 = 0 x = 10 x = 4

8 (iii) (iv) x x 10 x negative positive negative (v) x 10 (vi) Since the inequality is > 0, we choose the intervals on which x 4 positive and include endpoints that do not have an open circle. Solution: (4, 10] x 10 x 4 is