Algebra 2 help please! completing the square.?

short answer: most of the time you can divide the middle number by 2 and square that. Make the third number into that result.

Longer answer...

You need to put the equation in the form (x-a)^2 = b^2

then you have (x-a) = +/- b

x = a +/- b

(x-a)^2 expands to x^2 - 2ax + a^2

First, divide your original equation by the coefficient of x^2.

4x2 + 20x + 25 = 49 --> x2 + 5x + 25/4 = 49/4

Next, divide the coefficient of x by -2 to get a.

5 / -2 = -5/2

The constant term needs to be a^2. Add or subtract from both sides of the equation to make this true.

a^2 = 25/4, don't need to change anything in this example.

Take the squareroot of the right side to get b.

sqrt(49/4) = 7/2

x = -5/2 +/- 7/2

x = -12/2, 2/2

x = -6, 1

get the constants on one side

take half the coefficent of x and square it.

add to both sides

5)

-6

-3

9

add 9

x^2-6x+9=-1

(x-3)^2=-1

take square roots

x-3=+-i

x=3+-i

C)

same process for all of them

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