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anonymous

  • 5 years ago

solve by completing the square.

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  1. anonymous
    • 5 years ago
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    x\[x ^{2}+2x=24\]

  2. anonymous
    • 5 years ago
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    move all one side you get\[x^{2} +2x -24 =0\]

  3. anonymous
    • 5 years ago
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    now you factour(x+6)(x-4)

  4. anonymous
    • 5 years ago
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    x=-6 x=4

  5. anonymous
    • 5 years ago
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    \[2x ^{2}-4x-3=0\]

  6. anonymous
    • 5 years ago
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    this is quadratic equation form: [-b±√(b^2-4(a)(c))]/(2(a)) a=2 b=-4 c=-3 [-(-4)±√((-4)^2-(4*2*(-3))]/(2*2)) [(4)±√(16+24)]/(4) [(4)±2√(10)]/(4) x =1/2[±√(10)]

  7. anonymous
    • 5 years ago
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    The above method is not "completing the square"

  8. anonymous
    • 5 years ago
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    annette the second you want do complete square?

  9. anonymous
    • 5 years ago
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    Even in first one, it is splitting the middle term method and not completing the square

  10. anonymous
    • 5 years ago
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    Thank you Harkirat for remind me let do again , can you do second one it late I have to go 2/2=1 , 1^2 =1 (x+1)^2=24+1 (x+1)^2=25 x +1 = √(25) x = -1± 5 x=-6 or x= 4

  11. anonymous
    • 5 years ago
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    For the second we proceed as follows: \[2x ^{2}-4x-3=0\]

  12. anonymous
    • 5 years ago
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    First we take -3 to RHS as 3\[2x ^{2}-4x=3\]

  13. anonymous
    • 5 years ago
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    Next we divide both sides by 2 (to get rid of the 2 in front of x^2) So we get \[x ^{2}-2x=3/2\]

  14. anonymous
    • 5 years ago
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    Now to complete the square we add 1^2 to both sides \[x ^{2}-2x+1^{2}=3/2 + 1\]

  15. anonymous
    • 5 years ago
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    Now left side is a complete square as follows \[(x-1)^{2}= 5/2\] This gives \[x-1 = \pm \sqrt{5/2}\] Now taking -1 to RHS we get the final answer \[x = 1 \pm \sqrt{5/2}\]

  16. anonymous
    • 5 years ago
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    So the two roots/zeros of the given quadratic equation are \[1+\sqrt{5/2}\] and \[1-\sqrt{5/2}\]

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