1 / 12

Quadratic Functions

Quadratic Functions. Sum and Product of Roots. Purpose. Find sum/product of roots without knowing the actual value of x. Use the sum/product of roots to solve for other results. Derivation of Formula. Given α and β are roots of a quadratic equation, we can infer that

scott
Download Presentation

Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Quadratic Functions Sum and Product of Roots

  2. Purpose • Find sum/product of roots without knowing the actual value of x. • Use the sum/product of roots to solve for other results

  3. Derivation of Formula Given α and β are roots of a quadratic equation, we can infer that x1 = α and x2 = β , where α and β are just some numbers. For example, x2 – 5x + 6 = 0 has roots α and β means that α = 2 and β = 3

  4. However… Sometimes, we do not need the values of x to help us solve the problem. Knowing the relationship between the quadratic equation and its roots helps us save time

  5. Derivation of Formula ax2 + bx + c = 0 has roots α and β –(1) x2 + x + = 0 This tells us that: (x – α)(x – β) = 0 Expanding, we have: x2 – αx – βx + αβ = 0 x2– (α + β)x + αβ = 0 So, if we compare the above with (1): Result: αβ = , α + β =

  6. In other words… ax2 + bx + c = 0 has roots α and β x2– (α + β)x + αβ = 0 x2– (sum of roots)x + product of roots = 0

  7. Application – Example 1 (TB Pg 55) Hence, 2x2 + 6x – 3 = 0 has roots α and β, find the value of (a) Without finding α and β, we know the value for α + β and αβ α + β = – 3 and αβ = – 3/2

  8. Application – Example 1 (TB Pg 55) 2x2 + 6x – 3 = 0 has roots α and β, find the value of (b) We already know that: α + β = – 3 and αβ = – 3/2 Expanding (b), we have

  9. Application – Example 2 (TB Pg 55) x2 – 2x – 4 = 0 has roots α and β, find the value of (a) Step 1: α + β = 2 and αβ = – 4 To find (a), we use the algebraic property Rearranging:

  10. Application – Example 2 (TB Pg 55) Hence, α − β = x2 – 2x – 4 = 0 has roots α and β, find the value of (b) Step 1: α + β = 2 and αβ = – 4 To find (b), we use the algebraic property Rearranging:

  11. Application – Example 2 (TB Pg 55) x2 – 2x – 4 = 0 has roots α and β, find the value of (c) Step 1: α + β = 2 and αβ = – 4 To find (a), we use the algebraic property Rearranging:

  12. Application – Example 3 (TB Pg 56) 2x2 + 4x – 1 = 0 has roots α and β, form the equation with roots α2 and β2 Remember! x2– (sum of roots)x + product of roots = 0 x2– (α2 + β2)x + α2β2 = 0 So in order to get the equation, I must find α2 + β2 and α2β2 How?

More Related