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Diana.xL

  • one year ago

http://prntscr.com/8m4yb0

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  1. Diana.xL
    • one year ago
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    @ganeshie8

  2. Diana.xL
    • one year ago
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    @phi

  3. phi
    • one year ago
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    do you see the 5x term? take the 5 from that term, divide it by 2 \[ \frac{5}{2}\] then square it \[ \frac{25}{4} \] add that to both sides of the equation

  4. phi
    • one year ago
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    what do you get?

  5. Diana.xL
    • one year ago
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    my answer would be b?

  6. phi
    • one year ago
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    what do you get after adding 25/4 to both sides of the equation?

  7. phi
    • one year ago
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    you should get \[ x^2 +5x + \frac{25}{4}= -1 + \frac{25}{4}\]

  8. phi
    • one year ago
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    the left side is a perfect square (that is why you add 25/4) it is \[ \left(x+ \frac{5}{2}\right)^2 = -1 + \frac{25}{4} \] I would write the -1 as -4/4 \[ \left(x+ \frac{5}{2}\right)^2 = \frac{-4}{4} + \frac{25}{4} \]

  9. phi
    • one year ago
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    on the right side you have two fractions with the same denominator, so you can add the tops

  10. Diana.xL
    • one year ago
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    21/4

  11. phi
    • one year ago
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    yes \[ \left(x+ \frac{5}{2}\right)^2 = \frac{21}{4} \] if you take the square root (of both sides to keep it equal) we get rid of the square on the left side \[ \left(x+ \frac{5}{2}\right)= \pm \sqrt\frac{21}{4} \\ x+ \frac{5}{2}= \pm\frac{\sqrt{21}}{2} \] now add -5/2 to both sides

  12. phi
    • one year ago
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    you get \[ x= \frac{-5 \pm \sqrt{21}}{2} \]

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