Category Archives: Trigsted Chapter 1

1.4 Quadratic Equations

Quadratic Equation: Completing the Square

Example: Solve the quadratic equation by completing the square

2x^2+8x-9=0

Solution:

Original quadratic equation
2x^2+8x-9=0
Move the constant to the other side of the equation by adding 9 to both sides.
2x^2+8x-9=0
2x^2+8x-9+9=0+9
2x^2+8x=9
Make the coefficient of the squared term 1 by dividing both sides of the equation by 2.
2x^2+8x=9
{2x^2+8x}/2=9/2
{2x^2}/2+{8x}/2=9/2
x^2+4x=9/2
Add the number that completes the square to both sides of the equation.  The number that completes the square can be found by starting with the coefficient of the x term, dividing by 2 and then squaring.
(4/2)^2
(2)^2
4
x^2+4x=9/2
x^2+4x+4=9/2+4
The left hand side factors to be a perfect square
x^2+4x+4=9/2+4
(x+2)(x+2)=9/2+4
(x+2)^2=9/2+4
Simplify the right hand side by finding a common denominator and adding the fractions.
(x+2)^2=9/2+4
(x+2)^2=9/2+4/1
(x+2)^2=9/2+{4*2}/{1*2}
(x+2)^2=9/2+8/2
(x+2)^2=17/2
Solve the equation using the square root method.  Start by taking the square root of both sides to get rid of the square
(x+2)^2=17/2
sqrt{(x+2)^2}=sqrt{17/2}
x+2=pm sqrt{17/2}
Solve for x by subtracting 2 from both sides.
x+2=pm sqrt{17/2}
x+2-2=-2pm sqrt{17/2}
x=-2pm sqrt{17/2}
Rationalize the denominator by applying the radical to the numerator and denominator
x=-2pm sqrt{17/2}
x=-2pm sqrt{17}/sqrt{2}
Rationalize the denominator by multiplying by the fraction  sqrt{2}/sqrt{2}
x=-2pm sqrt{17}/sqrt{2}
x=-2pm sqrt{17}/sqrt{2} sqrt{2}/sqrt{2}
x=-2pm sqrt{17}/sqrt{2}
x=-2pm sqrt{34}/sqrt{4}
x=-2pm sqrt{34}/2

1.1 Linear Equations

Solving a Linear Equation with Faction Coefficients

Example:  Solve the equation.

1 /9 (6t-9)=11/9 t -{t+9}/18

Solution:

Solve the equation by clearing fractions.
1 /9 (6t-9)=11/9 t -{t+9}/18
The least common multiple of the denominators is 18.  Multiply the least common multiple by each term.
1 /9 (6t-9)=11/9 t -{t+9}/18
18*{1 /9} (6t-9)=18*{11/9} t -18*{t+9}/18
As you simplify, the fractions cancel.
18*{1 /9} (6t-9)=18*{11/9} t -18*{t+9}/18
2 (6t-9)=2  (11) t -(t+9)
Use the distributive property and combine like terms.
2 (6t-9)=2  (11) t -(t+9)
12t-18=22 t -t-9
12t-18=21 t -9
Move the variables to one side of the equation and simplify.
12t-18=21 t -9
12t-21t-18=21 t-21t -9
-9t-18=-9
Get the variable by itself on one side of the equation by adding 18 on both sides and dividing by -9.
-9t-18=-9
-9t-18+18=-9+18
-9t=9
{-9t}/-9=9/-9
t=-1

The solution to the equation is -1.

Here is a youtube video that might help as well.

1.5 Applications of Quadratic Equations, Applications of Quadratic Equations

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}.

1.2 Applications of Linear Equations, Applications of Linear Equations

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:

1.1 Linear Equations, Solving Equations

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:

1.1 Linear Equations, Solving Equations

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
1.6 Other Types of Equations, Equations, Equations that are Quadratic in Form

Quadratic in Form (U-substitution)

Example: Solve the equation.

5x^{2/3}-6x^{1/3}+1=0

Solution:

The equation is similar to a quadratic.  It has 3 terms and one exponent is twice the other.  Since the equation is quadratic in form, use substitution to solve the equation.

Use the following substitution to rewrite the equation

u=x^{1/3}

u^2=x^{2/3}

Original Equation
5x^{2/3}-6x^{1/3}+1=0
Substitute
u=x^{1/3}
u^2=x^{2/3}
5x^{2/3}-6x^{1/3}+1=0
5u^{2}-6u+1=0
Solve the quadratic equation by factoring.
1) Factor the quadratic
5u^{2}-6u+1=0
(5u-1)(u-1)=0
Solve the quadratic equation by factoring.
2) Apply the zero product property
(5u-1)(u-1)=0
5u-1=0 or u-1=0
Solve the quadratic equation by factoring.
3) Solve each linear factor
5u-1=0 or u-1=0
5u-1+1=0+1 or u-1+1=0+1
5u=1 or u=1
{5u}/5=1/5 or u=1
u=1/5 or u=1
Substitute again to bring back the original variable.  Use the original substitution.
u=x^{1/3}
u=1/5 or u=1
x^{1/3}=1/5 or x^{1/3}=1 
Solve the equation with rational exponents.
1) Rewrite the rational exponents in radical form
x^{1/3}=1/5 or x^{1/3}=1
root{3}{x}=1/5 or root{3}{x}=1
Solve the equation with rational exponents.
2) Cancel the cube root by cubing both sides.
3) Simplify
root{3}{x}=1/5 or root{3}{x}=1
(root{3}{x})^3=(1/5)^3 or (root{3}{x})^3=(1)^3
x=1/125 or x=1

The solution to 5x^{2/3}-6x^{1/3}+1=0 isx=1/125 or x=1.

 

Here is a video with similar examples.