SOLVING QUADRATICS Copyright - Kramzil Pty Ltd trading as Academic Teacher Resources
SOLVING QUADRATICS General Form: y a b c Where a, b and c are constants
To solve a quadratic equation, the equation must be epressed in the form: a b c That is, all variables and constants must be on one side of the equals sign with zero on the other.
Methods for Solving Quadratic equations a b c Method 1 : Factorisation Method : Quadratic Formula b b 4ac a Method 3 : Completing the Square Method 4 : Graphic Calculator
Method 1 : Factorisation Eample 1 : 1 Step 1: Equation in the right form 1 Step : Factorise 1 Step 3: Separate the two parts of the product either ( 1) or ( ) Step 4: Solve each equation 1 or
Some points to note Eample 1 : 1 1 1 either Need to remember your factorisation skills The equation is epressed as the product of two factors being equal to zero, therefore, one (or both) of the factors must be zero. 1 or 1 or To check your answers are correct you can substitute them one at a time into the original equation.
Checking solutions Eample 1 : 1 1 1 1(1) 11 1() 4 4
Method 1 : Factorisation Eample : 3 8 Step 1: Equation in the right form 3 8 Step : Factorise 7 4 Step 3: Separate the two parts of the product either 7 or 4 Step 4: Solve each equation 7 or 4
Method : Quadratic Formula Eample 1 : 8 5 Step 1: Determine the values of a, b and c a = 1 b = 8 and c = 5 Step : Substitute the values of a, b and c Step 3: Simplify This step is optional. Students need to have covered surds 8 ( 8) (1) 8 64 8 84 8 1 b 4(1)( 5) b 4ac a See net slide
Note that is a factor of the numerator 8 1 Once is factored out, it can be cancelled with the in the denominator 4 1 4 1 Eact answer 4 4 1 1 8.58.58 Decimal approimation
Method : Quadratic Formula Eample : 3 8 Step 1: Determine the values of a, b and c a = 3 b = 8 and c = Step : Substitute the values of a, b and c Step 3: Simplify This step is optional. Students need to have covered surds 8 8 (3) 8 64 4 6 8 4 6 8 6 1 b 4(3)() b 4ac a See net slide
Note that is a factor of the numerator 8 6 1 Once is factored out, it can be cancelled with the 6 in the denominator 4 6 1 4 3 1 Eact answer 4 3 4 3 1 1.8.39 Decimal approimation
Method : Quadratic Formula Eample 3 : 5 7 6 Step 1: Determine the values of a, b and c a = 5 b = 7 and c = 6 Step : Substitute the values of a, b and c Step 3: Simplify Problem - you cannot find the square root of a negative number 7 ( 7) (5) 7 491 1 7 71 1 b 4(5)(6) No real Solutions b 4ac a
That was a lot of work to find that there was no solution! It would be useful to be able to test the equation before we start. For this we use the DISCRIMINANT.
The Discriminant b 4ac The discriminant is a quick way to check how many real solutions eist for a given quadratic equation. As shown above the symbol for the discriminant is and it is calculated using b 4ac.
Summary of Results using Discriminant b 4ac > = < The equation has two real solutions The equation has one real solutions The equation has no real solutions
Relating the Discriminant to graphs b 4ac > y) y y The graph cuts the -ais in two places. These are the two real solutions to the quadratic equation.
Relating the Discriminant to graphs b 4ac = y) y y These graphs have their turning point on the -ais and hence there is only the one solution.
Relating the Discriminant to graphs b 4ac < y) y y There are no solutions in this case because the graphs do not intersect with the -ais.
Method 3: Completing the Square Technique Eample 1 : 6 3 This value is half b ( 3) ( 9 3) 3 6 Subtract the square of the number in the bracket Add 6 to both sides ( 3) 6 Take the square root of both sides 3 6 Subtract 3 from both sides 3 6
Method 3: Completing the Square Technique 3 6 This result gives us the eact answers. 3 6 or 3 6 Use your calculator to find decimal approimations accurate to two decimal places..55 or 5.45
Method 3: Completing the Square Technique Eample : This value is half b Add 57 4 ( to both sides Take the square root of both sides Subtract.5 from both sides 5 ) 5 ( 5 5 4 ) ( 8 8 57 4 5 ) 57 4.5 57 5 Subtract the square of the number in the bracket 57
Eample 3: Solve 7 1 [( 7 1 ( 3.5) 1.5 1 This value is half b ( 3.5) 3.5) ( (.5].5 3.5) 3.5) Factor out the coefficient of ² Subtract the square of the number in the bracket.5.5 See net slide
( 3.5).5 3.5.5 3.5.5 i.e. 3.5. 5 or 3.5. 5
Method 4 Graphic Calculator Graph the function and find the value of the -intercepts Use a solver function for a polynomial of degree