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anonymous

  • one year ago

A quadratic equation is shown below: 9x2 − 36x + 36 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points) Part B: Solve 2x2 − 9x + 7 = 0 using an appropriate method. Show the steps of your work and explain why you chose the method used. (5 points)

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  1. freckles
    • one year ago
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    So part A sounds like it i asking you to find the discriminant

  2. freckles
    • one year ago
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    \[ax^2+bx+c=0 \\ \text{ then the discriminant is } b^2-4ac\]

  3. anonymous
    • one year ago
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    I already have part A I just need Part B.

  4. freckles
    • one year ago
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    \[\text{ if } b^2-4ac>0 \text{ then you have two real solutions } \\ \text{ if } b^2-4ac=0 \text{ then you have one real solution (multiplicity 2)} \\ \text{ if } b^2-4ac<0 \text{ then you have two complex solutions }\] (assuming of course that a,b,c are real coefficients)

  5. welshfella
    • one year ago
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    for part B you can use factorisation

  6. freckles
    • one year ago
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    Well you can use quadratic formula. Or completing the square.

  7. freckles
    • one year ago
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    Or factoring.

  8. anonymous
    • one year ago
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    so it would be 1

  9. anonymous
    • one year ago
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    @freckles

  10. freckles
    • one year ago
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    that would be one solution

  11. freckles
    • one year ago
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    there is another solution

  12. freckles
    • one year ago
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    And we are talking about 2x^2-9x+7=0 right?

  13. anonymous
    • one year ago
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    yes

  14. freckles
    • one year ago
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    so if you figured out x=1 is solution then you know that x-1 is a factor (x-1)( )=2x^2-9x+7 what goes in then second pair of ( )

  15. freckles
    • one year ago
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    you know that x times what is equal to 2x^2? and that -1 times what is equal to 7?

  16. anonymous
    • one year ago
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    -7

  17. freckles
    • one year ago
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    yes but what about the other question x times what gives you 2x^2

  18. anonymous
    • one year ago
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  19. freckles
    • one year ago
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    \[x \cdot 2x=2x^2 \\ -1 \cdot (-7)=7 \\ \text{ and the middle term will be } -2x-7x \text{ which is indeed } -9x\] \[(x-1)(2x-7)=2x^2-9x+7 \\ \\ \text{ so you have that the equation } 2x^2-9x+7=0 \\ \text{ is equivalent to the equation } (x-1)(2x-7)=0 \\ \text{ which is equivalent to solving the conjunction } x-1=0 \text{ or } 2x-7=0\]

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