Category Archives: Trigsted Chapter 1

Application of Quadratic Equation: Translation

Example:  The product of some negative number and 5 less than twice that number is 273.  Find the number.

Solution:  Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

x  (expression that comes next)

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows.  Twice that number refers to 2 times the negative number that was described before.  (2x-5)

x  (2x-5)

“…is 273”

This translates to equals 273.

x (2x-5)=273

Solve the equation.

 

The translated equation
x (2x-5)=273
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)
2x^2-5x=273
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula)  I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.
2x^2-5x=273
2x^2-5x-273=273-273
2x^2-5x-273=0
Factor
2x^2-5x-273=0
(       )(       )=0
(2x       )(x       )=0
(2x      21)(x     13)=0
(2x    +  21)(x  -   13)=0
Use the zero product property and set each factor equal to zero and solve.
(2x    +  21)(x  -   13)=0
2x    +  21=0 or x  -   13=0
2x    +  21-21=0-21 or x  -   13+13=0+13
2x =-21 or x =13
{2x}/2 ={-21}/2 or x =13
x ={-21}/2 or x =13

Since the statement says that the number is negative, the number is {-21}/2.

Application of Linear Equation (Simple Interest)

Example:  Larry invested part of his $31,000 advance at 6% annual simple interest and the rest at 5% annual simple interest.  If his total yearly interest from both accounts was $1,760, find the amount invested at each rate.

Solution:  This question involved simple interest.  We will use the simple interest formula.

I=Prt

The I stands for interest, the P stands for principal (initial investment), r stands for the interest rate and t stands for time in years.

When starting an application problem like this it can be helpful to organize the information in a table using the formula.

I P r t
6% Account
5% Account

Begin filling in the information from the problem.  The interest rate is the most obvious and you should use the decimal version of the percent by moving the decimal place left two places.  Since the problem state that $1760 is the yearly interest for both accounts, we will use a time of 1 year.

I P r t
6% Account .06 1
5% Account .05 1

The goal in this question is to find the amount invested in each account.  I will let x be the amount invested in the 6% account.  The problem states that the rest of the $31,000 will be invested in the 5% account.  We can represent the rest of the money as 31000-x.

I P r t
6% Account x .06 1
5% Account  31000-x .05 1

You can fill in the interest column using the simple interest formula.

For the 6% account: I=Prt=(0.06)(x)(1)=0.06x

For the 5% account: I=Prt=(0.05)(31000-x)(1)=0.05(31000-x)

I P r t
6% Account 0 .06x x 0.06 1
5% Account 0.05(31000-x) 31000-x 0.05 1

You can now create an equation with this information.  The interest from the first account plus the interest from the second account should equal the total interest of $1760.

0.06x+0.05(31000-x)=1760

Solve the equation.
0.06x+0.05(31000-x)=1760
Use the distributive property and combine like terms to simplify each side of the equation.
0.06x+1550-0.05x=1760
0.01x+1550=1760
Solve for x by subtracting 1550 on both sides and simplifying.
0.01x+1550-1550=1760-1550
0.01x=210
Solve for x by dividing both sides by 0.01 and simplifying.
{0.01x}/0.01=210/0.01
x=21000

Since x represents the amount invested in the 6% account, $21000 is invested in the 6% account and the rest is invested in the 5% account.  The rest is $31,000-$21,000=$10,000.  $10,000 is invested in the 5% account.

Video Example:

Rational Equation (no solution)

Example: Solve the rational equation.

{x+4}/{x-3}+1=7/{x-3}

Solution:

{x+4}/{x-3}+1=7/{x-3}

Since we are solving a rational equation we need to first find the restrictions (the values of x that cause the expression to be undefined).

To find the restrictions create an equation by setting each denominator equal to zero and solving.

x-3=0

x-3+3=0+3

x=3

Having x=3 causes a zero in the denominator and the overall expression undefined.  That makes 3 a restricted value .

With the restriction in mind we will solve the equation.

 

The original equation
{x+4}/{x-3}+1=7/{x-3}
Multiply each side of the equation by the least common multiple of the denominators.  For this equation the least common multiple is x-3
(x-3)({x+4}/{x-3}+1)=(x-3)7/{x-3}
Distribute the least common multiple to each term.
(x-3){x+4}/{x-3}+1(x-3)=(x-3)7/{x-3}
Simplify by canceling the common factors.  This should clear any denominators.
x+4+1(x-3)=7
Use the distributive property to simplify.
x+4+x-3=7
Simplify each side of the equation by combining like terms.
2x+1=7
Solve for x by getting x by itself on one side.  Start by subtracting 1 on both sides.
2x+1-1=7-1
2x=6
 Solve for x by getting x by itself on one side.  Next divide both sides by 2.
{2x}/2=6/2
x=3
Compare your solution to the restricted value.
Since the solution is the same as the restricted value we must exclude it as a solution.  Since all of the solutions have been excluded, there is no solution to the rational equation.

Video Example:

Higher Order Equation that reduces to a linear equation

Example: Solve the equation.

(x+5)^3-9=x(x+7)(x+8)-6

Solution:

The original equation
(x+5)^3-9=x(x+7)(x+8)-6
Simplify both sides of the equation.  On the left hand side, rewrite the exponent.  On the right hand side, begin to simplify the multiplication.
(x+5)(x+5)(x+5)-9=x(x^2+8x+7x+56)-6
Simplify both sides of the equation.  On the left hand side, begin multiplying.  On the right hand side, combine like terms.
(x+5)(x^2+5x+5x+25)-9=x(x^2+15x+56)-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  On the right hand side use the distributive property.
(x+5)(x^2+10x+25)-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, continue multiplying.  The right hand side is in simplest form.
x(x^2+10x+25)+5(x^2+10x+25)-9=x^3+15x^2+56x-6
x^3+10x^2+25x+5x^2+50x+125-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  The right hand side is in simplest form.
x^3+15x^2+75x+116=x^3+15x^2+56x-6
Now that each side is in simplest form we want the terms with x on one side and the constant terms on the the other side.  Subtract x^3 from each side.  It cancels from each side.
x^3-x^3+15x^2+75x+116=x^3-x^3+15x^2+56x-6
15x^2+75x+116=15x^2+56x-6
Subtract 15x^2 from each side.  It cancels from each side.
15x^2-15x^2+75x+116=15x^2-15x^2+56x-6
75x+116=56x-6
Subtract 56x from each side and simplify. 
75x-56x+116=56x-56x-6
19x+116=-6
Subtract 116 from each side and simplify.
19x+116-116=-6-116
19x=-122
Get x by it self by dividing by 19 on both sides and simplify.
{19x}/19={-122}/19
x={-122}/19

Powers of i

Example: Simplify the power of i.

i^-59

Solution:

i^-59

Rewrite the exponential expression with negative exponents by using the reciprocal property.

=1/{i^59}

Simplify the i^59 by dividing 59 by 4.  The remainder will tell you which position in the pattern you are in.  59 divided by 4 is 14 with a reminder of 3.

=1/{i^3}

=1/{-i}

=1/{-i}  {i /i}

=i/{-i^2}

=i/{-(-1)}

=i/{1}

=i