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Thread: Factorisation and Completing the Square

  1. #1
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    Factorisation and Completing the Square

    Maths is certainly not me, why on earth I decided to pick Further is beyond comprehension. A couple of questions have got me, eternal thanks for any response.

    1. Solve quadratic by factorising:

    $\displaystyle y^2+8y=0$


    2. Solve by completing the square:

    $\displaystyle 3x^2-5x-1=0$


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Nick87 View Post
    1. Solve quadratic by factorising:

    $\displaystyle y^2+8y=0$
    What factor is common to both terms?


    Quote Originally Posted by Nick87 View Post
    2. Solve by completing the square:

    $\displaystyle 3x^2-5x-1=0$
    $\displaystyle 3x^2 - 5x - 1 = 0$

    $\displaystyle x^2 - \frac{5}{3}x - \frac{1}{3} = 0$

    $\displaystyle x^2 - \frac{5}{3}x = \frac{1}{3}$

    Now, you have this equation and the left hand side is in the form of
    $\displaystyle x^2 + 2bx$

    By comparison we get that $\displaystyle 2b = -\frac{5}{3} \implies b = -\frac{5}{6}$.

    We want to add something to $\displaystyle x^2 + 2bx$ that will make it a perfect square. If you look at the formula $\displaystyle (a + b)^2 = a^2 + 2ab + b^2$ you will see that if we add a $\displaystyle b^2$ to the two terms we get
    $\displaystyle x^2 + 2bx + b^2 = (x + b)^2$

    So we want to add a $\displaystyle b^2 = \frac{25}{36}$ to both sides of the equation:
    $\displaystyle x^2 - \frac{5}{3}x + \frac{25}{36} = \frac{1}{3} + \frac{25}{36}$

    So...

    $\displaystyle \left ( x - \frac{5}{6} \right ) ^2 = \frac{37}{36}$

    Can you finish it from here?

    -Dan
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  3. #3
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    Or for the second one:

    $\displaystyle \begin{aligned}3x^2 - 5x - 1 &=0\\ 36x^2 - 60x - 12 &=0\\ (6x - 5)^2 &= 37.\end{aligned}$

    Dan's answer is in agreement with mine.
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