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anonymous

  • 5 years ago

solve the equation: the square root of x^2-3x+25=x+2

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  1. anonymous
    • 5 years ago
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    set it =0 and use the quadratic formula.

  2. anonymous
    • 5 years ago
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    \left\{\left\{x\to 2-i \sqrt{19}\right\},\left\{x\to 2+i \sqrt{19}\right\}\right\}

  3. anonymous
    • 5 years ago
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    \[\left\{\left\{x\to 2-i \sqrt{19}\right\},\left\{x\to 2+i \sqrt{19}\right\}\right\}\]

  4. anonymous
    • 5 years ago
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    \[\sqrt{x ^{2}-3x+25}=x+2\]

  5. anonymous
    • 5 years ago
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    wow what are you doing? That doesn't look right at all.

  6. anonymous
    • 5 years ago
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    thats what the problem is , that s what i have to solve

  7. anonymous
    • 5 years ago
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    Do you know what the quadratic formula is?

  8. anonymous
    • 5 years ago
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    yeah but you dont use the quadratic formula for this problem

  9. anonymous
    • 5 years ago
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    you got to square both sides but after that i have no idea what to do

  10. anonymous
    • 5 years ago
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    Why can't you use the quadratic formula?

  11. anonymous
    • 5 years ago
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    because this is solving radical equations

  12. anonymous
    • 5 years ago
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    But this time you have another value of x aside from x^2

  13. anonymous
    • 5 years ago
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    \[(\sqrt{x ^{2}-3x+25})^{2}=(x+2)^{2}\]

  14. anonymous
    • 5 years ago
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    that problem is that you took a square root out of nowhere.

  15. anonymous
    • 5 years ago
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    This is the first step in solving it , but then i dont know what to do, this is how i have it in my notes and how it has it in the book

  16. anonymous
    • 5 years ago
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    a square cancels the square root on the left side

  17. anonymous
    • 5 years ago
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    But if you take the square root of one side you have to take the square root of the other side. So that squared thing is pointless.

  18. anonymous
    • 5 years ago
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    idk that s how the professor told us to do it

  19. anonymous
    • 5 years ago
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    Ok do that square root thing but do it to BOTH SIDES

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spraguer (Moderator)
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