Algebra: Chapter 13, Lesson 3, page 586 AND Chapter 13, Lesson 2, page 580 (because of yesterday’s snow day and the upcoming CST/STAR testing (ugh …).
Solving quadratics by completing the squares
We can use the technique of completing the square to solve quadratic equations. Recall that the addition property allows us to add a number to both sides of the equation. To complete a square, take `1/2` of the x-coefficient, square it and add it to both sides of the quadratic equation.
For example:
`x^2 – 4x – 7 = 0` becomes `x^2 – 4x = 7` by adding 7 to both sides.
Now, take `1/2` of the x coefficient (the `-4` in `-4x`). `1/2` of the `-4` is `- 2`. Now square that, `-2^2=4` and
`x^2 – 4x + 4 = 7 + 4` by adding 4 to both sides to complete the square.
REMEMBER: `x^2 – 4x + 4` factored becomes `(x – 2)^2`
`(x – 2)^2 = 11` or by square rooting each side to `x – 2 = ±sqrt(11)` so that
`x = 2 ± sqrt(11)` or `2 + sqrt(11)` and `2 – sqrt(11)` as the final solutions.
Purplemath explains it too, just click here for step-by-step instructions.
Two of tonight’s homework problems solved by MrE are here! Just click it.
Algebra: Chapter 13, Lesson 2, page 580.
More Solving Quadratic Equations
Solve a quadratic equation of the form `ax^2 = k`
Example: `-3x^2 + 7 = 0` becomes …
`-3x^2 = -7` or `x^2=7/3`
then `x = ±sqrt(7/3)`, don’t forget to rationalize this to `x = ±sqrt(21)/3`!
Solve a quadratic equation by factoring one expression into a binomial square [of the form `(x + a)^2 = k`]
Example: `(x-5)^2 = 9`, if you take the square root of both sides becomes …
`x-5 = ±sqrt(9)` or `x = 5 ±sqrt(9)` or `x = 5 ± 3` or `x=8` or `x=2`
Two of tonight’s homework problems solved by MrE are here! Just click it.
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Algebra 1a: CST Review – Equalities and Inequalities – GRAPHING
In graphing equalities and inequalities in x (number line) and x-y graphs, remember to use a T-chart for x-y graphs and to reverse the sign when graphing inequalities when dividing or multiplying by a negative number.
Remember too:
Definitions: `b` is the base and `h` is the height of the geometric figure below. Remember too, that the base and the height are at RIGHT ANGLES to each other!
Area, of a Triangle, `A = (b*h)/2` which is the same thing as `(1/2) * b * h`
Area of a Parallelogram, `A= b*h`.
Area of a Trapezoid, `A = (1/2) * h * (b_1 + b_2)` which is the same thing as `[h * (b_1 + b_2)]/2`