Algebra Tutorials! Saturday 16th of February

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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
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## Solving by extracting square roots

EXAMPLE:

Solve by extracting square roots

a: Solve 5x = 500

First divide both sides by 5! Then notice that the equation only involves two terms! Also notice that the value left on the right hand side of the equation is a perfect square number!

5x = 500 Given equation

x = 100 Divide by 5

Thus x = Â± 10 meaning x can be +10 or x can be -10!

b: Solve x = 7

Notice that the equation only involves two terms! Also notice that the value left on the right hand side of the equation is NOT a perfect square number! So what !!! You can still solve this by extracting square roots.

x = 7

So x = Â± 7

c: Solve (x- 2) = 64

METHOD 1: Notice that the left hand side of the equation is still a squared term. It just happens to be x -2! It can be solved by taking the square root of both sides as follows:

(x- 2)= 64

So x - 2 = Â±8

Thus x - 2 = 8 or x - 2 = -8

x = 10 or x = -6

METHOD 2: Clear the parentheses, collect like terms on left, set = 0, factor!

(x- 2) = 64

x -4x + 4 = 64

x -4x + 4 - 64 = 0

x -4x - 60 = 0

(x - 10)(x + 6) = 0

Either x - 10 = 0 or x + 6 = 0

x = 10 or x = -6