Discriminant, Applications, Hidden Quadratics

The Discriminant and Factoring

If we have a quadratic expression

    ax² + bx + c

and want to determine if it factors, then we can ask an equivalent question about whether there are rational roots.  In particular, there will be rational roots if the part under the square root sign in the quadratic formula is a perfect square.  We call this part the discriminant which has the form

    D  =  b² - 4ac 

Example

Determine if     

        3x² + 2x - 1     

can be factored.

Solution

We calculate

        D  =  2² - 4(3)(-1)  =  4 + 12  =  16

Since this is a perfect square, it can be factored.  In fact, it has the factorization

    3x² + 2x - 1   =  (3x -1)(x + 1)

Example

Determine if     

        5x² + 8x + 2     

can be factored.

Solution

We calculate

        D  =  8² - 4(5)(2)  =  64 - 40  =  24

Since this is not a perfect square, it can not be factored. 

Finding the Quadratic From its Roots

The zero factor property tells us that if r is a root of a quadratic, then x - r is a factor.  We can use this to reconstruct a quadratic equation from its roots.

Example

Find a quadratic equation that has roots 2 and -3.

Solution

Since 2 and -3 are roots, we write

        (x - 2) (x - (-3))  =  0

        (x - 2) (x + 3)  =  0

        x² + 3x - 2x - 6  =  0

        x² + x - 6  =  0

Example

Find a quadratic expression that has roots 2 - and 2 + .

Solution

We write

        (x - (2 - ))(x - (2 + ))  =  0

        =  ((x - 2) + )((x - 2) - )  =  0

        =  (x - 2)² - 3  =  0        Difference of Squares

        =  x² - 4x + 4 - 3  =  0

        =  x² - 4x + 1  =  0

Exercises

Find a quadratic equation that has roots

A.  -2, 5                            

B.  4 + , 4 -        

C.  1 + i, 1 - i                 

Hidden Quadratics:

Often we encounter an equation that is a quadratic in disguise.  For example the equation

        x⁴  + 3x²  -  4  =  0

is a fourth degree equation.  Notice that the exponent of the leading term is twice the exponent of the second term.  Because of this we can use what is called a substitution:

Let 

        u  =  x²  

then 

        x⁴  =  u²   

Then the equation in terms of u becomes:

        u²   + 3u - 4  =  0

which is a quadratic.  Hence, we can factor:

        (u - 1)(u + 4)  =  0

So that 

        u = 1     or     u = -4

Since u = x²  we have that

        x²  = 1     or     x²  = -4

The first has solution 

        x = 1     or     x = -1  

and the second has no solution.

Exercises

Solve

A.  x^2/3 - x^1/3 - 6 = 0                         

B.  (3x² - 4)² + (3x² - 4) - 2  =  0       

Applications

Some pointers on solving word problems.

1. Read the question and ask yourself what the problem is asking.
   
   

2. Label variables:  Write "Let x = ..."  where x should be a numerical variable that will give the solution.  Then label any other variables if needed.
   
   

3. If possible draw the picture. 
   
    Recall the Pythagorean Theorem:  
   
           a² + b² = c²  
   
   where a and b are the lengths of the legs of the right triangle and c is the length of the hypoteneuse.
   
   

4. Reread the problem phrase by phrase, inserting the variables when they are spoken in the phrase.  (Recall that is ,will be, was, are, equate to =).
   
   

5. Using the phrases come up with appropriate equations.
   
   

6. Solve the equations.
   
   

7. Reread the problem and answer it making sure that the answer makes sense.

Example

Avery throws a football straight up in the air.  The equation

        s  =   -16t² + 90t

gives the distance s in feet that the football is above the ground t seconds after he throws it.  

How high is the ball after 2 seconds?  after 30 seconds?

How long does it take for the ball to hit the ground?

Solution

Here we are given the variable so we do not need to label it.  First we calculate

    s(2)  =  -16(2)² + 90(2)  =  116

The ball is 116 feet above the ground after 2 seconds.

Next we calculate 

    s(30)  =  -16(30)² + 90(30)  =  -6300

Notice that the negative number implies that the ball has already hit the ground.  Hence the ball is on the ground after 30 seconds.  

For the last question, we know that when the ball hits the ground, it is 0 feet above the ground.  We solve

        0  =  -16t² + 90t  =  2t(-8t + 45)

Since t is not zero (this is when the player throws the ball), we have

        -8t + 45  =  0

        -8t  =  -45

        t  =  -45/8  =  5.625

It will take 5.625 seconds for the ball to hit the ground.

Example

Two ships leave the same port at the same time.  The first ship travels due North and the second travels due East.  After 1 hour the ships are 25 miles apart.  The ship traveling North travels 5mph faster than the ship traveling East.  How fast are the ships traveling?

Solution:

We form a right triangle with 25 the length of the hypotenuse.  

        Let x = the speed of the ship traveling East.  

        Then x + 5 is the speed of the ship traveling due North.  

Since 

        d  =  rt

the legs of the triangle are 

        (x)(1) 

and 

        (x + 5)(1)

The Pythagorean theorem tells us that

        x² + (x + 5)²   =  25²  

or 

        x² + x²  + 10x + 25  =  625.

        2x² + 10x - 600  =  0 

or 

        x² + 5x - 300  =  0

        (x -15)(x + 20)  =  0 

so that 

        x = 15      or      x = -20

Since x must be positive, x = 15.  

The ship traveling North is traveling at 20 mph and the ship traveling East is traveling at 15 mph.

Exercises

1. The product of two numbers is 85.  What are the numbers if one number is 2 more than three times the other?
   
   
   
   
   

2. One Leg of a right triangle is 12 feet and the hypotenuse is 3 feet longer then twice the other leg.  Find the length of the other leg.
   
   
   
   

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