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}

 

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

 

Application of Quadratic Function: Maximize

Example:

A baseball player swings and hits a pop fly straight up in the air to the catcher.  The height of the baseball in meters t seconds after it is hit is given by the quadratic function h(t)=-4.9t^2+17.4t+1.  How long does it take the baseball to reach its maximum height?  What is the maximum height obtained by the baseball?

Solution:

h(t)=-4.9t^2+14.7t+1 is defined to be a quadratic function.  The graph of a quadratic function is a parabola.  This quadratic function has a leading coefficient of -4.9 which means that the parabola is opening down.

Find the vertex to find the maximum value.

The formula for the x coordinate of the vertex
t=-b/{2a}
For a quadratic function a is the coefficient of the square term, b is the coefficient of the linear term, and c is the constant
h(t)=-4.9t^2+14.7t+1
a=-4.9; b=14.7; c=1
Substitute the values of a and b into the formula
t=-b/{2a}
t={-(14.7)}/{2(-4.9)}
Simplify with a calculator
t={-(14.7)}/{2(-4.9)}
t=1.5

The ball reaches the maximum height 1.5 seconds after the ball was hit.

The maximum height can be found by substituting 1.5 seconds for time in the height function.

The height function
h(t)=-4.9t^2+14.7t+1
Substitute t=1.5 in the height function
h(t)=-4.9t^2+14.7t+1
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
Simplify using a calculator
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
h(1.5)=12.025

The maximum height of the ball is 12.025 meters.

 

Form a Polynomial given the Degree and Zeros

Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 4; Zeros -2-3i; 5 multiplicity 2

Solution:

By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.

There are three given zeros of -2-3i, 5, 5.

The remaining zero can be found using the Conjugate Pairs Theorem.  f(x) is a polynomial with real coefficients.  Since -2-3i is a complex zero of f(x) the conjugate pair of -2+3i is also a zero of f(x).

Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.

Since f(x) has a zero of 5, f(x) has a factor of x-5

Since f(x) has a second zero of 5, f(x) has a second factor of x-5

Since f(x) has a factor of -2-3i, f(x) has a factor of x-(-2-3i)

Since f(x) has a factor of -2+3i, f(x) has a factor of x-(-2+3i)

Form the polynomial using all of the factors.  The leading coefficient will remain unknown.
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
Multiply the factors with complex numbers.  Doing so will cancel the complex numbers from the expression

  • Distribute the minus
  • Multiply each term in one factor by each term in the other factor
  • simplify i^2=-1
  • combine like terms
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
f(x)=a(x-5)^2(x+2+3i)(x+2-3i)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9i^2)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9(-1))
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i+9)
f(x)=a(x-5)^2(x^2+4x+13)
Multiply the other pair of factors
f(x)=a(x-5)^2(x^2+4x+13)
f(x)=a(x-5)(x-5)(x^2+4x+13)
f(x)=a(x^2-5x-5x+25)(x^2+4x+13)
f(x)=a(x^2-10x+25)(x^2+4x+13)
Multiply the two trinomials by multiplying each term in the first trinomial by each term in the other trinomial and then combine like terms
f(x)=a(x^2-10x+25)(x^2+4x+13)
f(x)=a(x^4+4x^3+13x^2-10x^3-40x^2-130x+25x^2+100x+325)
f(x)=a(x^4-6x^3-2x^2-30x+325)

The polynomial with degree 4 and zeros of -2-3i and 5 wiht multiplicity 2 is f(x)=a(x^4-6x^3-2x^2-30x+325)

Application: Exponential Growth and Decay (Half-life)

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution:  There is a two part process to this problem.  Part 1: Use some of the information to find the decay rate of radium.  Part 2: Answer the question using the rest of the given information.

Part 1:  Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

A=A _0 e^{kt}

A _0 is the initial amount

k is the decay rate

t is the time

A is the amount after t time has passed

Since radium has a half life of 1690 years, we know that whatever initial amount of radium is present after 1690 years there will be half of that initial amount left.  This allows me to identify A=1 /2 A _0 when the t=1690.

Substitute these values into the exponential decay formula and solve for k.

 

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for A and t since it is known that the half-life is 1690 years A=1 /2 A _0 and t=1690
A=A _0 e^{kt}
1 /2 A _0=A _0 e^{k(1690)}
Solve for the decay rate k:
Start by dividing both sides by the coefficient to isolate the exponential factor
1 /2 A _0=A _0 e^{k(1690)}
{1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0
1 /2= e^{k(1690)}
Solve for the decay rate k:
Take the natural log of both sides to get k out of the exponent
1 /2= e^{k(1690)}
ln(1 /2)= ln(e^{k(1690)})
Solve for the decay rate k:
Use the power rule for logarithms to get k out of the exponent
ln(1 /2)= ln(e^{k(1690)})
ln(1 /2)= k(1690)ln(e)
Solve for the decay rate k:
Simplify ln e = 1
ln(1 /2)= k(1690)ln(e)
ln(1 /2)= k(1690)(1)
ln(1 /2)= k(1690)
Solve for the decay rate k:
Solve for k by dividing by 1690 on both sides
ln(1 /2)= k(1690)
{ln(1 /2)}/1690= {k(1690)}/1690
{ln(1 /2)}/1690= k
-0.000410 approx k

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2:  Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years.  Symbolically that is A _0=50 when the t=630.  Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for initial amount (A_0), time (t), and the decay rate (k).
A _0=50 , t=630, and k={ln(1 /2)}/1690
A=A _0 e^{kt}
A=50 e^{630{ln(1 /2)}/1690}
Type the expression into your calculator and round to the thousandth place.
A=50 e^{630{ln(1 /2)}/1690}
A approx 38.615

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

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.

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