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Math Help - Factoring Quadratic Equations. Help needed...

  1. #1
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    Unhappy Factoring Quadratic Equations. Help needed...

    Would someone please help me understand the last few steps of this problem... I've gotten this far but do not know where to go from where I left off...
    I know the answer is y=5/4, y=3/2

    Factoring Quadratic Equations. Help needed...-img_3159.jpg

    Thank you so much in advance
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  2. #2
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    Re: Factoring Quadratic Equations. Help needed...

    read the section on completing the square

    Pauls Online Notes : Algebra - Quadratic Equations - Part II
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  3. #3
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    Re: Factoring Quadratic Equations. Help needed...

    Hello, smyatsallie!


    \begin{array}{c} 8y^2 - 22y + 15 \;=\;0 \\ \\ y^2 - \frac{11}{4}y + \frac{15}{8} \;=\;0 \\ \\ y^2 - \frac{11}{4}y \;=\;-\frac{15}{8} \\ \\ y^2 - \frac{11}{4}y + \frac{121}{64} \;=\;-\frac{15}{8} + \frac{121}{64} \\ \\ \left(y - \frac{11}{8}\right)^2 \;=\;\frac{1}{64} \end{array}

    Take the square root of both sides:

    . . y - \tfrac{11}{8} \;=\;\pm\sqrt{\tfrac{1}{64}}

    . . y - \tfrac{11}{8} \;=\;\pm\tfrac{1}{8} \quad\Rightarrow\quad y \;=\;\tfrac{11}{8} \pm \tfrac{1}{8}

    Therefore: . \begin{Bmatrix}y &=& \frac{11}{8} + \frac{1}{8} &=& \frac{12}{8} &=& \frac{3}{2} \\ \\ y &=& \frac{11}{8} - \frac{1}{8} &=& \frac{10}{8} &=& \frac{5}{4} \end{Bmatrix}
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  4. #4
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    Re: Factoring Quadratic Equations. Help needed...

    Thank you so much. So much easier to understand when broken down like that. I appreciate it.
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