x 3 = ± 7 x = 3 ± 7 E.g. Solve the equation: (2x + 5) 2 = 23 Ans: (2x + 5) 2 = 23 2x + 5 = ± 23 x = 5 ± 23

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1 If we have the equation x = 1, what are the solutions to the equation? Since the square of a positive or negative numer are always positive, this equation has two solutions, namely x = 4 or x = 4. What aout the equation x = 10? 10 is not a perfect square, ut using radicals, we can come up with two possile solutions too: x = 10 or x = 10 In general, all equations of the form x = a, where a is a positive numer, will always give two solutions, namely x = a or x = a. We use the symo ± to mean plus or minus. So we will write x = ± a. Solve the equation: x = 7 Ans: x = ± 7 Solve the equation: x = 35 Ans: x = ± 35 Solve the equation: x = 300 Ans: x = ± 300 = ± = ±10 3 Notice that if x = and is a negative numer, then we have no (real numer) solution, since the square of a real numer is always positive. Example: Solve: x = 5 Ans: The equation has no (real numer) solution. In using the property of square root to solve an equation of the form x = a, the expression eing squared does not have to e just x, ut can e more complicated form: Solve the equation: (x 3) = 7 Ans: (x 3) is the expression eing squared (it is our new x), following a similar example, we have: (x 3) = 7 1

2 x 3 = ± 7 x = 3 ± 7 Solve the equation: (x + 5) = 3 Ans: (x + 5) = 3 x + 5 = ± 3 x = 5 ± 3 x = 5 ± 3 Solve the equation: (4x + 7) = 8 Ans: (4x + 7) = 8 4x + 7 = ± 8 = ± 4x = 7 ± x = 7 ± 4 Completing the Square Some quadratic equations cannot e solved y factoring. For example, x 8x+1 = 0 cannot e solved y factoring ecause the left hand side cannot e factored. Is there another way to try to solve this equation? We know from the property of radicals that if an equation is of the form x = a, where a is a positive numer, then x = ± a are the two solutions to the equation. Is there a way to turn x 8x + 3 into a form that is a perfect square? If we look at (x + ) = (x + )(x + ) = x + x +, we see that a quadratic expression is a perfect square if the constant term (the term) is half of the middle term (the term) squared. This gives us a way to turn a quadratic expression into a perfect square: Make sure that the coefficient on x is 1. Take half of the coefficient

3 of the x term, square the result, and use that as the constant term. x 8x + 3 = 0 We first move the constant of 1 to the right side, since we need a new constant term: x 8x = 3 We take half of the coefficient of x, that is, we take half of 8, we get 4. We square this numer, ( 4) = 1. This is our new constant term. We add this term to oth sides of the equation: x 8x + 1 = x 8x + 1 = 13 The left hand side is now a perfect square, we can factor it: (x 4) = 13 The property of radical/squares tells us that x 4 = ± 13 Solving for x gives: x = 4 ± 13 In other words, we get two solutions to the given equation, they are: x = or x = 4 13 This method is called The Method of Completing the Square. What if the coefficient of x is not 1? We would then divide oth sides of the equation y the appropriate constant to make the coefficient of x to e 1 first: 3x + x 3 = 0 For this equation, the coefficient of x is 3, not 1. In order to apply the method of completing the square, we divide oth sides of the equation 3

4 y 3 first: 3x + x 3 3 = 0 3 x x 1 = 0 We now apply the same method again. Note that this time, fractions will e involved. x x = 1 We take half of 1 3, which is 1, and square this numer, which gives 1 3. This is the new constant we want to use: x x = x x = 37 3 Factor the left hand side gives: ( x + 1 ) = 37 3 x + 1 = ± 37 3 x + 1 = ± 37 x = 1 ± 37 Unlike the factoring method, the method of completing the square works on any quadratic equations, and will always give us any/all solution to a quadratic equation with real coefficients. Quadratic Formula: The method of completing the square can e used to solve any quadratic equation with real coefficient. The only prolem with it, though, is that it is rather tedious to apply. Instead, we apply the method of completing the square once on a generic quadratic equation to come up with a formula: 4

5 General quadratic equation: ax + x + c = 0 We first divide oth sides of the equation y a to make the coefficient of x to e 1. ax + x + c = 0 a a x + a x + c a = 0 x + a x = c a Taking half of a gives, and we add the square of this to oth sides a of the equation, this gives: x + ( a x + = c ( a a + a The left hand side is now a perfect square and can e factored: ( x + = c ( a a + a x + a = ± c ( a + a x = a ± c ( a + a Simplify the right hand side gives us the Quadratic Formula: x = ± 4ac a The quadrtic formula is a wrapped up version of the method of completing the square. It can e used to solve any quadratic equation just like the method of completing square. It is, however, more efficient than completing the square since it only gives the answer. In the quadratic formula, the expression inside the radical, 4ac, is called the discriminent, and its value determines the numer of real 5

6 numer solutions the quadratic equation has: If 4ac > 0, the quadratic equation has real numer solutions. If 4ac = 0, the quadratic equation has 1 real numer solution. If 4ac < 0, the quadratic equation has no real numer solution. In applying the quadratic formula, make sure that you recognize what are the coefficients a,, and c. x + 3x 5 = 0 In this equation, a = 1, = 3, c = 5 x = 3 ± (3) 4(1)( 5) (1) = 3 ± 9 x x + 1 = 0 In this equation, a =, =, c = 1 x = ( ) ± ( ) 4( )(1) ( ) 3x 4x 5 = 0 = ± 1 4 = 1 ± 3 In this equation, a = 3, = 4, c = 5 x = ( 4) ± ( 4) 4(3)( 5) (3) x x + 13 = 10 = 4 ± 7 = 3 ± 19 3 We cannot try to use the quadratic formula on this equation at the moment. Instead, make the right hand side 0 first: x x + 3 = 0

7 Now, a = 1, = 1, c = 3 x = ( 1) ± ( 1) 4(1)(3) = 1 ± 11 Note that the numer inside the radical is a negative numer, 11. Since the radical of a negative numer is not a real numer, we say that the aove equation have no real numer solution. While the quadratic formula (or completing square) can e used to solve any quadratic equation, in the case that an expression can e factored, it is still much faster to use the factoring method. Therefore, in trying to solve a quadratic equation, it is still a good idea to try to solve y factoring first. Use quadratic formula only if you cannot factor the expression. 7