ALGEBRA A Topic Overview A SOLUTION OF QUADRATIC EQUATIONS This topic describes three methods of solving Quadratic equations. assumes you understand and have practised using the algebraic methods described in Topics A and A (Lessons A. and A. on Factorisation are particularly relevant). LESSON PLAN Lesson No. LESSON TITLE Page Examples SAQs A. Factorisation Method Factorisation Differences of squares, Perfect squares. -.. -. A. A. A. Formula Method The Quadratic Roots formula The Discriminant 6. A. A. A. Graphical Method.6 -.7 A. Appendix A. Completing the Square Method.8 Appendix A. Solutions to the Self-Assessment Questions Page
A Quadratic Equations A QUADRATIC EQUATIONS Quadratic equations involve the second power of x, the unknown quantity. The general quadratic equation has the form ax + bx + c (a ) Quadratic equations are solved to give two possible solutions for x. i.e. two possible roots. Quadratic equations can be solved by:. Factorisation Method. Quadratic Formula Method. Graphical Method Page
A. Factorisation Method A. FACTORISATION METHOD As described in the A. and A. on factorisation, a quadratic expression can sometimes be written as the product of two linear factors. EXAMPLE. Solve the quadratic equation by factorisation: x + x + Solution: First, factorise the equation x + x + (x + )(x + ) For this equation to be true the LHS must equal zero, then either x + or x + x or x Therefore, x, x are the solutions or the roots of the equation. CHECK: Substitute the roots back into the original quadratic equation. x + x + If x LHS ( ) + ( ) + + RHS If x LHS ( ) + ( ) + 8 + RHS The solutions are correct. Page
A. Factorisation Method EXAMPLE. Solve the equation by factorisation: x + 7x Solution: First, factorise the equation x + 7x (x )(x + ) For this equation to be true the LHS must equal zero, then either x or x + x or x x. or x Therefore, x., x are the solutions or the roots of the equation. CHECK: Substitute the roots back into the original quadratic equation. x + 7x If x. LHS (.) + 7(.). +. RHS If x LHS ( ) + 7( ) RHS The solutions are correct. SAQ A. Solve the following quadratic equations by factorisation: (a) x x + (b) x x (c) x + 7x (d) x + x Page
A. Factorisation Method There are two cases which need to be recognised when solving quadratic equations. EXAMPLE. Differences of Squares Solve the equation: x Solution: x (x + )(x ) x + or x x or x Note: This is the difference of two squares Therefore, the roots are x or x The general solution for this type of equation is: x k (x + k)(x k) x + k or x k x k or x k Therefore, the roots are x k or x k EXAMPLE. Perfect Squares Solve the equation: x + 6x + Solution: x + 6x + (x + )(x + ) x + or x + x or x Therefore, the roots are x (twice) Both the roots are the same and, in these cases, the root is known as a double root. The general solution for this type of equation is: x + kx + k (x + k)(x + k) (x + k) x k (twice) SAQ A. Solve the following quadratic equations by factorisation: (a) x (b) x 8 (c) x + 8x + 6 (d) x (e) x + x + (f) x Page
A. Formula Method A. QUADRATIC EQUATIONS BY FORMULA It is not always possible to factorise a quadratic equation, but the following formula can be used to solve all quadratic equations of the form: ax + bx + c b (b ac) x ± a The following examples illustrate how this formula is used to solve quadratic equations. EXAMPLE. Solve the following equations using the quadratic formula: (i) x x + (ii) x + 6x Solution: (i) To solve x x + First, identify a, b and c in the quadratic equation. In this example: a, the coefficient of x b, c +, the coefficient of x the constant Then, substitute these values into the formula giving: Hence ( ) ± x ± ± ± 6 x or x or ( ) () (6 ) ()() Therefore, the solutions of the equation are x or x In this case we can observe that the solutions imply that x and x, so it follows that (x )(x ) We can thus see that x x + (x )(x ) so the formula method can provide the factors of the original expression. Page 6
A. Formula Method CHECK: x x + If x LHS () () + + RHS If x LHS () () + + RHS The solutions are correct. (ii) To solve x + 6x First, identify a, b and c in the quadratic equation. In this example: a, the coefficient of x b 6, c, the coefficient of x the constant Then, substitute these values into the formula giving: 6 ± (6) ()( ) x () 6 ± (6 + ) 6 ± 76 6 ± 8.78 6 + 8. 78 Hence, x or x 6 8. 78.78.78 Therefore, the solutions of the equation (rounded to decimal places accuracy) are x.7 or x.7 CHECK: x + 6x If x.7 LHS (.7) + 6(.7). If x.7 LHS (.7) + 6(.7). Note: You are unlikely to get exactly zero when substituting rounded solutions into the LHS. To obtain more accurate solutions, you will have to work to more decimal places, eg. substituting x.78 gives LHS.76 Page 7
A. Formula Method Note: A relationship exists between the roots and the coefficients a, b, c of a quadratic equation ax + bx + c such that Sum of the Roots Product of the Roots b a c a Looking back at the first example, where x x + with roots x, x Sum of the Roots and b ( ) a Product of the Roots and c a In the following SAQ, check that the above relationships hold for your solutions, as well as checking the solutions by substituting into the quadratic equations. SAQ A. Use the quadratic formula to solve these equations and check the solutions: (a) x x (b) x + x + (c) m m + (d) 7t 8t + (e) y y (f) x + x 6 (g) + x + x Page 8
A. Formula Method The Discriminant: We are now going to focus on the term b ac in the quadratic formula. It is called the Discriminant. There are three cases to consider: CASE : b ac Consider the equation x + 6x + (x + )(x + ) Hence x + or x + x or x In this case, both roots have the value Let us now consider the discriminant b ac where a, b 6 and c Then b ac (6) ()() 6 6 In general, if b ac, then the roots are equal and real. CASE : b ac > Consider the equation x x + (x )(x ) Hence x or x x or x In this case, the roots are distinct and real. Let us now consider the discriminant b ac where a, b and c Then b ac ( ) ()() 6 > In general, if b ac >, then there are two distinct real roots. CASE : b ac < Consider the equation: x + x + As the equation does not factorise, let us use the formula to solve the equation b ± x ± ± (b a () ( ) ac) ()() where a, b, c Note: We cannot find the square root of a negative number. In this case b ac < In general, if b ac <, then the roots are not real. Page
A. Formula Method Summarising the three cases, comparing the value of the Discriminant: b ac > Roots are real and distinct b ac Roots are equal and real b ac < Roots are not real SAQ A. Without solving the equations, find the nature of the roots of the following quadratic equations: (a) x + x + (b) x + x + 7 (c) x + x + (d) x 8x + 8 (e) y + 6y + (f) c c + Page
A. Graphical Method A. GRAPHICAL METHOD The graphical method usually starts with plotting pairs of values (x, y) on graph paper. A graph is then drawn by sketching a curve through the plotted points. The roots of a quadratic equation are determined from the x values of the points at which the graph crosses the x-axis. A step by step description of the graphical method of solving quadratic equations is given in the following examples. EXAMPLE.6 Find the roots of the equation x + 6x + by the Graphical Method. Solution: Step : Let y x + 6x + Step : Construct a table of values of y x + 6x + corresponding to suitably selected values of x. TABLE: Tabulation of y x + 6x + x 6 x 6 6 6x 6 8 6 6 y x + 6x + Step : On graph paper, plot the values of y for each value of x. Then sketch a curve through the plotted points to produce a graph which will look like: Graph of y x + 6x + 8 6 y x -6 - - - - -6 Page
A. Graphical Method Step : Use the graph to find the roots of the equation y x + 6x + Observe the values of x at the points where the graph crosses the x-axis. (ie. where y x + 6x + ) We can see that: y when x and x Thus the roots of x + 6x + are x and x Note: We can verify the relationship between the roots and the coefficients of the above quadratic equation: b 6 We expect: Sum of roots 6 a and we find: Sum of roots + ( ) 6 We also expect: Product of roots c a and we find: Product of roots ( )( ) EXAMPLE.7 Find the roots of the equation x x by the Graphical Method. Solution: Step : Let y x x Step : Construct a table of values of y x x corresponding to suitably selected values of x. TABLE: Tabulation of y x x x x x + +8 + 8 x 6 6 6 6 y x x Page
A. Graphical Method Step : Plot the points (x, y) and sketch the graph of y x x y x - - - - - - - - - Step : From the graph, observe that y when x. and x. So the roots of x x are x. and x. SAQ A. Use the graphical method to solve: (a) x x + (b) 6x + x Page
A. Completing the Square Method A. COMPLETING THE SQUARE METHOD Expressions such as x, (x + ), (x + c) are called perfect squares. x x ± Similarly (x + ) x + ± x ± EXAMPLE.8 Solve the equation x + 6x by 'completing the square'. Solution: Collect all terms to the LHS side with the coefficient of x positive. x 6x + Divide throughout by to make the coefficient of x equal to x x + (i) Half the coefficient of x giving (ii) Square the result giving ( ) (iii) Add the square result in (ii) above to both sides x x + + x x + + Now we can make a perfect square: x x + x Hence x + Subtract from both sides x 7 x Take the square root of both sides x ± 7 giving the solution x ± 7 Page
A. APPENDIX A. SOLUTIONS TO SELF ASSESSMENT QUESTIONS SAQ A. Factorisation Method (a) x x + (b) x x (x )(x ) (x + )(x ) x or x x + or x x x x x Roots are x, x Roots are x, x (c) x + 7x (d) x + x (x + )(x ) (x 7)(x + ) x + or x x 7 or x + x x x 7 x Roots are x, x Roots are x, x SAQ A. Differences of Squares/Perfect Squares (a) x (b) x 8 (x )(x + ) (x )(x + ) x ± x ± (c) x + 8x + 6 (d) x (x + ) x x (twice) (x )(x + ) x ± (e) x + x + (f) x (x + 7) x x 7 (twice) (x )(x + ) x ± Page
A. APPENDIX SAQ A. Formula Method (a) x x a, b, c ( ) ± x ± ± 6 x or x or x., x ()( ) (b) x + x + a, b, c x ± ( )( ) ± ± x or x 8 x, x (c) m m + a, b, c ( ) ± ()() m () ± 6 6 ± 8 6 8 m or m 6 m, m 6 Page 6
A. APPENDIX (d) 7t 8t + a 7, b 8, c 8 ± t 8 ± 6 8 ± 6 t or t t, t 7 6 ( 7)( ) (e) y y Rearrange the equation y y + a, b, c ± y ( )() ± 6 ± 8 6 y or y y, y (f) x x + 6 The equation has to be rearranged into standard form. Remove the fractions by multiplying each term by 6x(x + ) 6x(x + ) 6x(x + ) 6x(x + ) x x + 6 6(x + ) 6x x(x + ) 6x + 6 6x x x + x + x 6 Solve: x + x 6 a, b, c 6 ± x ()( 6) ± ± 6 x or x x, x Page 7
A. APPENDIX (g) + x + x Remove the fractions by multiplying each term by (x + )(x ) (x + )(x )() (x + )(x ) + (x + )(x ) x + x (x ) + (x + ) x x + x + x x x + x x + Solve: x x + a, b, c ± x 6 ()() ± ±.6 7.6.6 x or x x.7 x.7 (rounded to decimal places) SAQ A. The Discriminant (a) x + x + a, b, c b ac 6 Since b ac >, Two roots are real and distinct. (b) x + x + 7 a, b, c 7 b ac 8 Since b ac <, Two roots are not real. (c) x + x + a, b, c b ac Since b ac, Roots are equal and real. (d) x 8x + 8 a, b 8, c 8 b ac Since b ac, Roots are equal and real. (e) y + 6y + a, b 6, c b ac 6 6 Since b ac, Roots are equal and real. (f) c c + a, b, c b ac 6 Since b ac <, Two roots are not real. Page 8
A. APPENDIX SAQ A. Graphical Method (a) x x + Step : Let y x x + Step : Tabulate y x x + x x x 6 x y x x + 8 Step : Draw the graph of y x x + y x - - - Step : Observe the points where the graph crosses the x-axis (ie. where y ) The roots are x. and x. (approx.) Page
A. APPENDIX (b) 6x + x Step : Let y 6x + x Step : Tabulate y 6x + x x x 6x 6 6 6 x 6 6 y 6x + x 6 6 Step : Draw the graph of y 6x + x y - - x - - -6-8 Step : Observe the points where the graph crosses the x-axis (ie. where y ) The roots are x.7 and x. Page