Applications, Radicals, and Hidden Quadratics

 

  1. 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 1:  

    One number is three times another.  The sum of their reciprocals is 20/3.  Find the numbers.

     

    1. Read the question.  

    2. Let     x = one number,         y = another number

    3. Not applicable

    4. x is three times y.  The sum of 1/x and 1/y is 20/3.

    5. x = 3y                1/x + 1/y = 20/3

    6. 1/(3y) + 1/y = 20/3

      Multiply by 3y:  1 + 3 = 20y

      y = 1/5         so         x = 3/5

    7. The first number is 3/5 and the other number is 1/5.

    Example 2:  

    Chris is 4 years older than Gina. In 6 years the sum of their ages will be 68.  what are their ages now? 

    1. Let         x = Chris' age             y = Gina's age

    2. NA

    3. x is four more than y.  The sum of six more than x and six more than y will be 68.  what are x and y?

    4.  x = 4 + y         (x + 6) + (y + 6) = 68

    5. [(4 + y) + 6] + (y + 6) = 68 

      2y + 16 = 68         2y = 52

      y = 26 

      x = 4 + y = 30

    6. Chris is 30 years old and Gina is 26 years old

     

    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 20 feet and the hypoteneuse is 30 feet.  Find the length of the other leg.

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

              -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?

  1. Radical Equations

    If we have an equation with a single radical then we follow the procedure:

    1. Isolate the radical so that the radical is alone on the left side of the equation with everything else on the other side of the equation.

    2. Square both sides of the equation

    3. Bring everything back to the left side of the equation so that the right side is a 0.

    4. Solve the resulting equation by factoring, the root method, the quadratic equation, of some other method.

    5. Plug your answer back into the original equation making sure that your result is an actual solution and not an extraneous one.

     

    Example:  

           



    2. 7x + 4 = (x + 2)2

       7x + 4 = x2 + 4x + 4

    3. x2 - 3x = 0

    4. (x)(x - 3) = 0

      x = 0 or x = 3

    5. ok

     

     

    Exercise

    Solve 

    Equations with two radicals:

    If an equation has two radicals, we follow the following procedure:

     

    1. Put the two radicals on the left hand side and everything else on the right hand side.

    2. Square both sides (be careful to FOIL the two radicals)

    3. This will result in an equation with one radical.  Now follow the procedures for one radical.

     

     

    Example:  

            Solve    











    5. (3x + 4)(x) = 4x2  

      3x2  + 4x = 4x2  

      x2  - 4x = 0

      (x)(x - 4) = 0

      x = 0 or x = 4

    6. Check:       - 2 - 0 = 0     ok

      - 2 - = 0     ok



  1. 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.

    Exercise

    x2/3 - x1/3 - 6 = 0  

    Hint let u = x1/3

 

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