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Quadratic Functions - The Discriminant: Solving for x

Learn what the discriminant is in quadratic functions and how it can be used to determine the number of solutions for x. Understand the significance of a positive, negative, or zero discriminant and its impact on the roots of the equation.

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Quadratic Functions - The Discriminant: Solving for x

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

  2. In  What can we say about ... To get a solution for x ? 25 = +5 or -5 1 = +1 or -1 (92) = +9 or -9 (-4) = can’t do

  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 ax2+bx+c=0

  5. Roots 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 ax2+bx+c=0

  6. b2-4ac > 0 Which is which? b2-4ac < 0 b2-4ac = 0 b2-4ac = 0 b2-4ac > 0 b2-4ac < 0

  7. Using the discriminant If b2-4ac > 0 Equation has two distinct roots. If b2-4ac < 0 Equation has no real roots. If b2-4ac = 0 Equation has repeated roots.

  8. How it is used - example Calculate the discriminant of 2x2+7x+7=0 and hence prove 2x2+7x+7 is always > 0 b2 - 4ac = 72 – (4 x 2 x 7) = 49 - 56 = -7 a = [coefficient of x2] b = [coefficient of x] c= [constant] = 2 = 7 = 7 If b2-4ac < 0 Equation has no real roots. Therefore, doesn’t cross the x-axis and is always positive

  9. If b2-4ac > 0 Equation has two distinct roots. How it is used - example For what values of ‘k’ does the equation 2x2 -3x+k=0 have real roots a = [coefficient of x2] b = [coefficient of x] c= [constant] = 2 = -3 = k b2 - 4ac > 0 (-3)2 – (4 x 2 x k) > 0 9 – 8k >0 9 > 8k 9/8 > k k < 9/8

  10. If b2-4ac < 0 Equation has no distinct roots. Have a go For what values of ‘k’ does the equation 3x2 + 5x+k=0 have no real roots a = [coefficient of x2] b = [coefficient of x] c= [constant] = 3 = 5 = k b2 - 4ac < 0 52 – (4 x 3 x k) < 0 25 – 12k < 0 25 < 12k 25/12 < k k > 25/12

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