Solving Quadratic Equations

Date: 2/15/96 at 15:4:3
From: Llano Net Internet Software
Subject: quadratic equations

I am very confused on how to do these problems. 
Please help me.

My first two questions are:

(3x-1) to the second power = 1 divided by 4

Instructions: Solve each equation by using the 
square root property.

(x-3) to the second power = -20

This has the same instructions.
_________________________________________

My second question is 

2x to the second power - 5x + 2 = 0

Instructions:

Solve each equation by factoring.
_________________________________________

My next question is:

(2+ -3)(2 + 1) = 5

Same instructions
______________________

My last question is 

2x to the second power + 3x - 2 = 0

Instructions: Solve each equation using the 
quadratic formula.

Date: 4/29/96 at 10:45:35
From: Doctor Steven
Subject: Re: quadratic equations

#1) (3x-1)^2 = 1/4

    then

    (3x-1) = Sqrt(1/4) = 1/2 or -1/2

    so

    3x-1 = 1/2 or -1/2.

    Add one to both sides and get:

    3x = 3/2 or 1/2

    Divide by three on both sides to get:

    x = 1/2 or 1/6

#2) (x-3)^2 = -20

    then

    x-3 = Sqrt(-20) = 2Sqrt(-5) or -2Sqrt(-5).

    Add three to both sides and get:

    x = 3 + 2Sqrt(-5) or 3 - 2Sqrt(-5).

#3) 2x^2 - 5x + 2 = 0.

    Then we know we have two factors that look something like 
    (?x + ?)* (?x + ?). We can figure out what the question marks 
    are by looking at the constants in front of the x's.  

    Note the constant at the end of the equation is 2.  How can we 
    get 2 by multiplying two numbers together.  The answers are 2*1 
    and -2*-1.  So our possibilities are:

         (?x + 2)*(?x + 1)      or
         (?x - 2)*(?x - 1).

    We want the middle number to be negative so we can rule out 
    the first choice, so we know (?x - 2)*(?x - 1). 

    We also know we want the x^2 term to be multiplied by 2. 
    How can we get this?  Well again the answers are 2*1 or -2*-1.
    Since we try to keep negatives out of our highest term we 
    throw the negatives out and say it must be 2*1.  So now the 
    question is where does the 2 go and where does the 1 go.  
    We can check to see which one it is:

    (2x -2)*(x - 1) = 2x^2 - 3x + 2,
    (x - 2)*(2x - 1) = 2x^2 - 5x + 2.

    So the factorization of 2x^2 - 5x + 2 = 0 is (x - 2)*(2x - 1) = 0.

    The only way for a product to be zero is if one of the 
    multipliers is zero, so if:

    1) x - 2 = 0,      or
    2) 2x - 1 = 0,

    then their product will be zero (i.e, 2x^2 - 5x + 2 will be 
    zero).

    So the solutions to this are x = 2, and x = 1/2.

#4) In this one I'm assuming you meant (2x + -3)(2x + 1) = 5.

    We get by using FOIL on the left side:

    4x^2 - 4x - 3 = 5.

    Subtracting 5 from both sides we get:

    4x^2 - 4x - 8 = 0.

    Dividing by four we get:

    x^2 - x - 2 = 0

    Factor this using the method above and obtain:

    (x - 2)*(x + 1) = 0.

    So the solutions are:

    x - 2 = 0, and
    x + 1 = 0.

    So x = 2, or -1.

#5) 2x^2 + 3x - 2 = 0.

    The quadratic equation states that for a quadratic equation of 
    the form:

     ax^2 + bx + c = 0

    its solutions are:

     x = [-b + Sqrt( b^2 - 4ac )]/2a    and

     x = [-b - Sqrt( b^2 - 4ac )]/2a.

    In this case a = 2, b = 3, c = -2.  Plug them in and get the 
    solutions as:

    x = [-3 + Sqrt( 9 - 4*2*-2 )]/4

    x = [-3 - Sqrt( 9 - 4*2*-2 )/4

    Simplify to get:

    x = [-3 + Sqrt(25)]/4
    x = [-3 - Sqrt(25)]/4.

    Simplify some more to get:

    x = 2/4 = 1/2

    x = -8/4 = -2.

    So the roots of this are x = 1/2, and x = -2.

Hope this helps.

-Doctor Steven,  The Math Forum