Algebra placement test, help!?

Question

I have been studing for an Algebra placement test that I will be taking on Wednesday. I have done really well, until last night. I came to the quadratic equation part of my study sheets and I cannot for the life of me remember how do complete the problems. Please help!

Answer 1

Send me your problems at [e-mail]

Answer 2

There are 3 ways to solve quadratic equations.

Some will factor. For example,
2x² + 5x - 3 = 0
(2x - 1)(x + 3) = 0
then set each factor to 0 and solve:
2x - 1 = 0
2x = 1
x = 1/2
x + 3 = 0
x = -3

Some are easy to solve by completing the square. Best if number in front of x² is 1 and number in front of x is even:
x² - 6x + 2 = 0
x² - 6x = -2
x² - 6x + 9 = -2 + 9
the 9 is (-6/2)²
(x - 3)² = 7
x - 3 = ±√7
x = 3 ± √7

but all quadratics can be solved using the quadratic formula: if ax² + bx + c = 0, then
x = -b/(2a) ± √(b² - 4ac) / 2a

3x² + 4x - 2 = 0
x = -4/6 ± √(16 - 4(-6)) / 6
x = -2/3 ± √40 / 6
x = -2/3 ± 2√10 / 6
x = -2/3 ± √10 / 3
it's actually more work to simplify the answer than to fill in the formula.

Answer 3

Textbooks often say there are three ways to solve a quadratic equation, namely, factoring, completing the square, and by the quadratic formula. The last method is easiest to use. The formula is this: to solve ax^2 + bx + c , with a not 0, take x = (-b +/- sqrt(b^2 - 4ac))/2a .

Furthermore, the discriminant is D = b^2 - 4ac. If D > 0, there are two real and distinct roots; if D = 0, there are two real and equal roots; and if D < 0, there are two complex roots which are conjugates.

Answer 4

by the way....a way to remember the quadratic formula is the following story : A negative boy (-b) couldn't decide (+ or -) whether or not to go to a radical party (sq. root of). The boy was squared and he missed out on 4 awesome chics (b square minus 4ac). The party was all over at 2am (all divided by 2a).

Answer 5

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Answer 6

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