How do I solve these?

Question

4x² - 4x - 4 = 0 by completing the square

x^6 - 64 = 0 by anything

9x⁴ - 24x³ + 16x² = 0 by anything

.01x² - .05x - .04 = 0 by quadratic formula

y = (x+1) (x-3) by graphing

determine the following if it is perpendicular, parallel, or neither (explain to me please)

L1: (-2,-1) (1,5)

L2: (1,3) (5,-15)

what is extracting the square roots?

Answer 1

The first question is easy basically divide by the "x" number.

The quadratic formula is

x=(b+/- the square root of b^2-4ac)/2a

To graph you need a graphing medium but to help you need a "x" and a "y".

A square root is a "perfect multiple" like 2 is the square root of 4, but 5 is NOT a square root of 26.

Hopes this helps.

Answer 2

1

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4x^2 - 4x - 4 = 0

4x^2 -4x -4 +1 -1 = 0

(4x^2 -4x + 1) -4 -1 = 0

(2x-1)^2 -5 = 0

(2x-1)^2 = 5

2x-1 = +/- sqrt(5)

2x = 1 +/- sqrt(5)

x = (1 +/- sqrt(5))/2

So:

x = (1+sqrt(5))/2 =  1.61803398874989484820

x = (1-sqrt(5))/2 = -0.61803398874989484820

#2

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x^6 - 64 = 0

x^6 = 64

x = +/- (64))^(1/6)  = +/- 2

So x = 2 or x = -2

#3

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9x^4 - 24x^3 + 16x^2 = 0

x^2 ( 9x^2 - 24x + 16) = 0

x^2 ( 3x - 4)(3x - 4) = 0

So:

x^2 = 0

x=0

or

3x - 4 = 0

3x = 4

x = 4/3

Therefore, x=0 or x = 4/3

#4

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0.01x^2 - 0.05x - 0.04 = 0

x^2 - 5x - 4 = 0

x = (-(-5) +/- sqrt((-5)^2 - 4*1*(-4))) / (2*1)

x = (5 +/- sqrt(25 + 16)) / 2

x = (5 +/- sqrt(41)) / 2

So:

x = (5 + sqrt(41))/2 = 5.70156211871642434324

or

x = (5 - sqrt(41))/2 = -0.70156211871642434324

#5

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slope 1 = rise/run = (-1-5)/(-2-1) = -6/-3 = 2

slope 2 = rise/run = (3-(-15))/(1-5) = 18/-4 = -9/2

These lines are neither parallel nor perpendicular.

If they were parallel, then their slopes would be equal.

If they were perpendicular, then they would have to

be negative reciprocals of each other.  Algebraically:

slope1 = -1/slope2