Algebra: Chapter 13, Lesson 3, page 586.
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.
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Algebra 1a: CST Review – Area of a Parallelogram, Triangle and Trapezoid
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`