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SOLVING QUADRATICS

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General Form: Where a, b and c are constants

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To solve a quadratic equation, the equation must be expressed in the form: That is, all variables and constants must be on one side of the equals sign with zero on the other.

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Methods for Solving Quadratic equations Method 2 : Quadratic Formula Method 1 : Factorisation Method 4 : Graphic Calculator Method 3 : Completing the Square

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Method 1 : Factorisation Example 1 : or Step 1: Equation in the right form Step 2: Factorise Step 3: Separate the two parts of the product Step 4: Solve each equation

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Example 1 : or Some points to note Need to remember your factorisation skills The equation is expressed as the product of two factors being equal to zero, therefore, one (or both) of the factors must be zero. To check your answers are correct you can substitute them one at a time into the original equation.

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Example 1 : Checking solutions

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Method 1 : Factorisation Example 2 : or Step 1: Equation in the right form Step 2: Factorise Step 3: Separate the two parts of the product Step 4: Solve each equation

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Method 2 : Quadratic Formula Example 1 : a = 1 b = – 8 and c = – 5 Step 1: Determine the values of a, b and c Step 2: Substitute the values of a, b and c Step 3: Simplify See next slide This step is optional. Students need to have covered surds

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Exact answer Decimal approximation Note that 2 is a factor of the numerator Once 2 is factored out, it can be cancelled with the 2 in the denominator

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Method 2 : Quadratic Formula Example 2 : a = 3 b = 8 and c = 2 Step 1: Determine the values of a, b and c Step 2: Substitute the values of a, b and c Step 3: Simplify See next slide This step is optional. Students need to have covered surds

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Exact answer Decimal approximation Note that 2 is a factor of the numerator Once 2 is factored out, it can be cancelled with the 6 in the denominator

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Method 2 : Quadratic Formula Example 3 : a = 5 b = – 7 and c = 6 Step 1: Determine the values of a, b and c Step 2: Substitute the values of a, b and c Step 3: Simplify Problem - you cannot find the square root of a negative number 2 Imaginary Solutions

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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.

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The Discriminant The discriminant is a quick way to check how many real solutions exist for a given quadratic equation. As shown above the symbol for the discriminant is and it is calculated using.

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Summary of Results using Discriminant > 0 The equation has two real solutions = 0 The equation has one real solutions < 0 The equation has no real solutions

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Relating the Discriminant to graphs > 0 The graph cuts the x -axis in two places. These are the 2 real solutions to the quadratic equation.

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Relating the Discriminant to graphs = 0 These graphs have their turning point on the x -axis and hence there are 2 equal solutions.

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Relating the Discriminant to graphs < 0 There are 2 imaginary solutions in this case because the graphs do not intersect with the x -axis.

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Method 3: Completing the Square Technique Example 1 : This value is half b Subtract the square of the number in the bracket Add 6 to both sides Take the square root of both sides Subtract 3 from both sides

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Method 3: Completing the Square Technique This result gives us the exact answers. Use your calculator to find decimal approximations accurate to two decimal places.

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Method 3: Completing the Square Technique Example 2 : Take the square root of both sides Subtract 2.5 from both sides This value is half b Subtract the square of the number in the bracket Add to both sides

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Example 3: Solve Factor out the coefficient of x² This value is half b Subtract the square of the number in the bracket See next slide

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Method 4 Graphic Calculator Graph the function and find the value of the x -intercepts Use a solver function for a polynomial of degree 2

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Quadratic Theory Introduction This algebraic expression: x 2 + 2x + 1 is a polynomial of degree 2 Expressions like this in which the highest power of x.

Quadratic Theory Introduction This algebraic expression: x 2 + 2x + 1 is a polynomial of degree 2 Expressions like this in which the highest power of x.

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