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Quadratic Functions (4) What is the discriminant What is the discriminant Using the discriminant Using the discriminant.

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Presentation on theme: "Quadratic Functions (4) What is the discriminant What is the discriminant Using the discriminant Using the discriminant."— Presentation transcript:

1 Quadratic Functions (4) What is the discriminant What is the discriminant Using the discriminant Using the discriminant

2  25 = +5 or -5  1 = +1 or -1  (9 2 ) = +9 or -9  (-4) = can’t do  In What can we say about... To get a solution for x ?

3  In What can we say about... If it’s negative then it has no solutions ---> cannot square root a negative number If it’s zero then it only has only solution

4 The discriminant This is the discriminant of the equation ax 2 +bx+c=0

5 Using the discriminant The discriminant can be used to give us important information about the roots of our quadratic. The “roots” are basically our solutions when ax 2 +bx+c=0 Roots

6 Which is which? b 2 -4ac > 0 b 2 -4ac < 0 b 2 -4ac = 0 b 2 -4ac > 0 b 2 -4ac < 0 b 2 -4ac = 0

7 Using the discriminant If b 2 -4ac > 0Equation has two distinct roots. If b 2 -4ac < 0Equation has no real roots. If b 2 -4ac = 0Equation has repeated roots.

8 How it is used - example Calculate the discriminant of 2x 2 +7x+7=0 and hence prove 2x 2 +7x+7 is always > 0 a = [coefficient of x 2 ] b = [coefficient of x] c= [constant] = 2 = 7 b 2 - 4ac = 7 2 – (4 x 2 x 7) = 49 - 56 = -7 If b 2 -4ac < 0Equation has no real roots. Therefore, doesn’t cross the x-axis and is always positive

9 How it is used - example If b 2 -4ac > 0Equation has two distinct roots. For what values of ‘k’ does the equation 2x 2 -3x+k=0 have real roots a = [coefficient of x 2 ] b = [coefficient of x] c= [constant] = 2 = -3 = k b 2 - 4ac > 0 (-3) 2 – (4 x 2 x k) > 0 9 – 8k >0 9 > 8k 9/8 > k k < 9/8

10 Have a go If b 2 -4ac < 0Equation has no distinct roots. For what values of ‘k’ does the equation 3x 2 + 5x+k=0 have no real roots a = [coefficient of x 2 ] b = [coefficient of x] c= [constant] = 3 = 5 = k b 2 - 4ac < 0 5 2 – (4 x 3 x k) < 0 25 – 12k < 0 25 < 12k 25/12 < k k > 25/12


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