Theory of Quadratic Equations PDF

MTH 101: ELEMENTARY MATHEMATICS I LECTURER: Prof O. M. Ogunlaran Mathematics Programme, College of Agriculture, Engineering and Science, Bowen University, Iwo, Nigeria Email...

MTH 101: ELEMENTARY MATHEMATICS I LECTURER: Prof O. M. Ogunlaran Mathematics Programme, College of Agriculture, Engineering and Science, Bowen University, Iwo, Nigeria Email Address: [email protected] 1 Theory of Quadratic Equations 1.1 Introduction The general form of a quadratic equation is: ax2 + bx + c = 0, (1) where a, b, and c are constants, and a ̸= 0. The solutions to the quadratic equation, i.e., the values of x that satisfy equation (1) are called the roots of the equation. There are four major techniques for solving quadratic equations: Factorization, Completing the Square, Quadratic Formula and Graphical Method. 1.2 Completing the Square The completing the square method is a technique used to solve quadratic equations by transforming the given quadratic expression into a perfect square trinomial. This method works for any quadratic equation of the form (1). Steps for Completing the Square Step 1: Rearrange the equation (1) by moving the constant term (c) to the right-hand side: ax2 + bx = −c. Step 2: Divide the entire equation by a (if a ̸= 1) to simplify the quadratic term: b c x2 + x = −. a a b Step 3: Take half of the coefficient of x (which is a ), square it, and add it to both sides of the equation.  2  2 2 b b c b x + x+ =− + a 2a a 2a Step 4: Rewrite the LHS as a squared binomial and simplify the RHS 2 b2 − 4ac  b x+ = 2a 4a2 Step 5: Take the square root of both sides of the equation: r b b2 − 4ac x+ =± 2a 4a2 1  Step 6: Solve for x: √ −b ± b2 − 4ac x= , (2) 2a which is called the Quadratic formula. Example Solving 2x2 + 8x − 6 = 0 by Completing the Square. Rearranging the equation, 2x2 + 8x = 6. Dividing by 2 (the coefficient of x2 ), x2 + 4x = 3. Adding the square of half the coefficient of x to both sides, x2 + 4x + 4 = 3 + 4 (x + 2)2 = 7 Taking the square root of both sides √ x+2=± 7 Solving for x √ x = −2 ± 7 √ √ Thus, the two solutions are x = −2 + 7 and x = −2 − 7. 1.3 Sum and Product of Roots The sum and product of the roots are helpful in various applications, such as: 1. Constructing a quadratic equation when the roots are known. 2. Checking the correctness of solved roots. 3. Solving problems in algebra where information about the sum and product of the roots is given instead of the actual roots. If the two roots of the quadratic equation (1) are denoted as r1 and r2 , then from (2) √ √ −b + b2 − 4ac −b − b2 − 4ac r1 = , r2 =. 2a 2a Equation (1) is equivalent to: (x − r1 )(x − r2 ) = 0 Expanding, we obtain x2 − (r1 + r2 )x + r1 r2 = 0. (3) Comparing equation (3) and b c x2 + x + = 0, a a b c we have that: r1 + r2 = − and r1 r2 =. a a Therefore, for any quadratic equation (1) with roots r1 , r2 , b c Sum of the roots: r1 + r2 = − , Product of the roots: r1 r2 =. a a 2 Examples Example 1: If the roots of 2x2 + 5x − 3 = 0 are r1 and r2 , find r1 + r2 and r1 r2. Comparing the given equation with the standard form (1), a = 2, b = 5 and c = −3. 5 3 Hence r1 + r2 = − , and r1 r2 = −. 2 2 √ √ Example 2: Construct an equation with roots 3 − 2, 3 + 2. √ √ √ √ √ In this case let r1 = 3 − 2 and r2 = 3 + 2, such that r1 + r2 = 3 − 2 + ( 3 + 2) = 2 3 and √ √ r1 r2 = ( 3 − 2)( 3 + 2) = −1. √ Hence the equation is x2 − 2 3x − 1 = 0. 1.4 Symmetric Functions of the Roots The values of other functions of r1 and r2 can easily be calculated provided they are symmetric and the value of r1 + r2 and r1 r2 are known for a given equation. For a symmetric function of r1 and r2 , the function remains the same or is multiplied by −1 if r1 and r2 are interchanged. For example, r12 + r22 is a symmetric function and so also r12 − r22 since it becomes −(r22 − r12 ). However, r12 + 2r22 is not symmetric. The values of a symmetric function of r1 and r2 can be obtained without knowing the values of the roots r1 and r2. Example 1: If r1 and r2 are the roots of 3x2 − 10x − 4 = 0, find the values of (a) r12 + r22 (b) r1 − r2 (c) (r1 − r2 )2. b 10 c 4 (a) Here a = 3, b = −10, c = −4. r1 + r2 = − = , r1 r2 = = −. a 3 a 3 Now,  10 2  4  124 r12 + r22 = (r1 + r2 )2 − 2r1 r2 = −2 − =. 3 3 9  10 2  4  148 (b) r1 − r2 cannot be found directly. So we use (r1 − r2 )2 = (r1 + r2 )2 − 4r1 r2 = −4 − =. √ √  3 3 9 148  2 37 Hence r1 − r2 = ± = depending on whether r1 > r2 or r1 < r2. 3 3  10  √148   20√37  2 (c) (r1 − r2 ) = (r1 + r2 )(r1 − r2 ) = = taking r1 > r2. 3 3 9 Example 2: If r1 and r2 are the roots of 3x2 + 5x − 1 = 0, construct equations whose roots are (a) r12 , r22 (b) α1 , β1. From the given equation, a = 3, b = 5, c = −1. (a) 25 2 31 r12 + r22 = (r1 + r2 )2 − 2r1 r2 = + =. 9 3 9 1 r12 r22 = (r1 r2 )2 =. 9 Therefore, the new equation is 31 1 x2 − x + = 0 9 9 2 or 9x − 31x + 1 = 0. (b) 1 1 r2 + r1 −5 + = = 13 = 5. r1 r2 r1 r2 −3  1  1   1  = = −3. r1 r2 r1 r2 3 Therefore, the new equation is x2 − 5x − 3 = 0. 1.5 Types of Roots The discriminant (∆) of the quadratic equation (1) helps determine the nature of the roots. The discriminant is defined as: ∆ = b2 − 4ac The value of the discriminant determines the number and type of roots (solutions) of the quadratic equation. There are three possible types of roots based on the value of ∆. 1. Two Distinct Real Roots (∆ > 0) If the discriminant is positive (∆ > 0), the quadratic equation has two distinct real roots. This happens when the value under the square root in the quadratic formula is positive, allowing two real solutions. The roots can be found using the quadratic formula: √ −b ± b2 − 4ac x= 2a Example 1: Consider the quadratic equation x2 − 5x + 6 = 0. Here, a = 1, b = −5, and c = 6. The discriminant is ∆ = (−5)2 − 4(1)(6) = 25 − 24 = 1. Since ∆ > 0, the equation has two distinct real roots. Solving the equation: √ −(−5) ± 1 5±1 x= = 2(1) 2 The two roots are: 5+1 5−1 x= = 3 and x = =2 2 2 Thus, the roots are x = 3 and x = 2. 2. Two Equal Real Roots (∆ = 0) If the discriminant is zero (∆ = 0), the quadratic equation has two equal real roots, also known as a repeated root or double root. This happens when the value under the square root in the quadratic formula is zero, giving only one solution. Example 2: Consider the quadratic equation x2 − 4x + 4 = 0. Here, a = 1, b = −4, and c = 4. The discriminant is ∆ = (−4)2 − 4(1)(4) = 16 − 16 = 0. Since ∆ = 0, the equation has two equal real roots. Solving the equation: √ −(−4) ± 0 4±0 x= = 2(1) 2 Thus, the only solution is: 4 x= =2 2 The root is x = 2 (a double root). 3. Two Complex Roots (∆ < 0) If the discriminant is negative (∆ < 0), the quadratic equation has two complex conjugate roots. In this case, the value under the square root is negative, and the roots are not real numbers. They involve √ imaginary numbers and are expressed in terms of i, where i = −1. The roots are of the form: p −b ± i |∆| x= 2a 4  Example 3: Consider the quadratic equation x2 + 4x + 5 = 0. Here, a = 1, b = 4, and c = 5. The discriminant is ∆ = (4)2 − 4(1)(5) = 16 − 20 = −4. Since ∆ < 0, the equation has two complex conjugate roots. Solving the equation: √ −4 ± −4 −4 ± 2i x= = 2(1) 2 The two roots are: −4 + 2i −4 − 2i x= = −2 + i and x = = −2 − i 2 2 Thus, the roots are x = −2 + i and x = −2 − i, which are complex conjugates. 5