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wilks311

  • 2 years ago

(x-5)(x+3)=33 ; find the real roots

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  1. cwrw238
    • 2 years ago
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    (x-5)(x+3)=33 x^2 - 2x - 15 = 33 x^2 - 2x - 48 = 0 can u continue from here?

  2. wilks311
    • 2 years ago
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    Thanks! Just realized my mistake. added 33 instead of subtracting

  3. cwrw238
    • 2 years ago
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    yw

  4. wilks311
    • 2 years ago
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    Ah another question. Havent done this stuff in 4 years.

  5. wilks311
    • 2 years ago
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    (x+2)^2(x-1)+(x+2)(x-1)^2

  6. whpalmer4
    • 2 years ago
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    \[(x+2)^2(x-1) + (x+2)(x-1)^2\]multiply it out, step by step: \[(x+2)^2 = (x+2)(x+2) = x^2+2x+2x+4 = x^2+4x+4\] \[(x^2+2x+4)(x-1) = x^3-x^2+2x^2-2x+4x-4 = x^3+x^2+2x-4\] \[x^3+x^2+2x-4 + (x+2)(x-1)^2\]you do the rest, then simplify

  7. wilks311
    • 2 years ago
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    I've gotten down to 2x^3+3x^2-11x-2 Any help factoring that please?

  8. whpalmer4
    • 2 years ago
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    @cwrw238 pretty sure that those 2s outside parentheses are supposed to be ^2

  9. cwrw238
    • 2 years ago
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    yea - u r right

  10. cwrw238
    • 2 years ago
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    i'm embarrassed!!

  11. whpalmer4
    • 2 years ago
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    life would be easier if openstudiers couldn't copy and paste the questions, losing all of those valuable positional clues :-)

  12. cwrw238
    • 2 years ago
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    yea

  13. wilks311
    • 2 years ago
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    Guys , you didnt need to expand!

  14. whpalmer4
    • 2 years ago
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    we didn't need to do anything, really :-)

  15. wilks311
    • 2 years ago
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    just do something similar like a^2v +ab^2 = ab(a+b) then solve!

  16. wilks311
    • 2 years ago
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    gah im hitting myself >:(

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