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Thursday, March 14, 2013

Solving Radical Equations, 2

Category: Algebra

"Published in Newark, California, USA"

Find the roots of the equation for



Solution:

In solving the radical equations, you have to be very careful and write down the equations very well because the roots of the radical equations can be real, extraneous, or both. Anyway, consider the given equation above



Transpose either one of the term to the right side of the equation



Divide both sides of the equation by their common factor which is (2x - 1) as follows



Square the both sides of the equation to eliminate the radical signs





Expand the equation above and solve for the value of x as follows









Next, substitute x = 0 to the given equation, we have

















Since the root is not satisfied to the given equation because  both sides of the equation are not equal, then the root of the equation is extraneous. NO ANSWER for this problem.


Wednesday, March 13, 2013

Simple Chemical Conversion

Category: Chemical Engineering Math, Differential Equations

"Published in Newark, California, USA"

For a substance C, the time rate of conversion is proportional to the square of the amount x of unconverted substance. Let k be the numerical value of the constant of proportionality and let the amount of unconverted substance be x0 at time t = 0. Determine x for all t ≥ 0.

Solution:

The given word problem is another good application of first order, first linear equation where a simple chemical conversion of a substance is involved. Let's analyze the given problem as follows

for                           C ────────────> Product

initial                       x0                                     0
at time t                 x0 - x                                  x

From the word problem, "the time rate of conversion is proportional to the square of the amount x of unconverted substance", the working equation will be



The above equation is negative because the value of x is decreasing as time is increasing. Using the method of separation of variables, we have



Integrate on both sides of the equation





To solve for the value of C, substitute x = x0 and t = 0 to the above equation, we have




Therefore, the amount of x at t ≥ 0 will be










Tuesday, March 12, 2013

Proving Trigonometric Identities, 2

Category: Trigonometry

"Published in Newark, California, USA"

Prove that



Solution:

In proving the trigonometric identities, we have to simplify the hardest or more complicated part of the equation. In this case, we have to simplify the left side of the given equation. Let's consider the following procedure.



but

The above equation becomes




















but


The above equation becomes







Therefore,