Week 6: Solving Quadratic Equations

Published by on March 12, 2018 | Leave a response

* General form: ax^2 +bx +c = 0 (a \neq 0)
There are three ways to solve quadratic equations:
– Factoring
– Using the Quadratic Formula:x=\frac{-b\pm\sqrt{b^2 -4ac}}{2a}
– Making perfect square
*To check if a trinomial can be factored, calculate \sqrt{b^2 -4ac}.
– If the result is a natural number, that expression can be factored.
– If the result is greater than 0 and not a natural number, use quadratic formula or make perfect square to solve the equation.
– If the result equals to 0, there is only 1 value of x.
– If the result is smaller than 0, there is no real value of x (extraneous solution).

* Example:
x^2 + x - 56 = 0
(\sqrt{1^2 - 4(1)(-56)} -> \sqrt{1 + 224} -> \sqrt{225} -> 15 -> can be factored)
– Factoring:
(x +8)(x – 7) = 0
x + 8 = 0 or x – 7 = 0
x = -8 or x = 7
– Using quadratic formula:
x = \frac{-1\pm15}{2(1)}
x = \frac{14}{2} or x = \frac{-16}{2}
x = 7 or x = -8

2x^2 + 9x + 8 = 0
(\sqrt{9^2 - 4(2)(8)} -> \sqrt{81 - 64} -> \sqrt{17} -> cannot be factored)
– Using quadratic formula:
x = \frac{-9\pm\sqrt{17}}{2(2)}
x = \frac{-9\pm\sqrt{17}}{4}
– Making perfect square:
x^2 + \frac{9x}{2} + 4 = 0
x^2 + 2x(\frac{9}{4}) + (\frac{9}{4})^2 - (\frac{9}{4})^2 + 4 = 0
(x + \frac{9}{4})^2 - \frac{81}{16} + 4 = 0
(x + \frac{9}{4})^2 - \frac{17}{16} = 0
(x + \frac{9}{4})^2 = \frac{17}{16}
x + \frac{9}{4} = \pm\frac{\sqrt{17}}{4}
x = \frac{-9 \pm \sqrt{17}}{4}

2x(x – 3) = 4(x – 3) + 1
2x^2 - 6x = 4x - 12 + 1
2x^2 - 6x - 4x + 12 - 1 = 0
2x^2 - 10x + 11 = 0
(\sqrt{(-10)^2 - 4(2)(11)} -> \sqrt{100 - 88} -> \sqrt{12} -> cannot be factored)
– Using quadratic formula:
x = \frac{-(-10)\pm\sqrt{12}}{2(2)}
x = \frac{10 \pm 2\sqrt{3}}{4}
x = \frac{5 \pm \sqrt{3}}{2}
– Making perfect square:
x^2 - 5x + \frac{11}{2} = 0
x^2 - 2x(\frac{5}{2}) + (\frac{5}{2})^2 - (\frac{5}{2})^2 + \frac{11}{2} = 0
(x - \frac{5}{2})^2 - \frac{3}{4} = 0
(x - \frac{5}{2})^2 = \frac{3}{4}
x - \frac{5}{2} = \pm\frac{\sqrt{3}}{2}
x = \frac{5 \pm \sqrt{3}}{2}

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