Expert: Paul Klarreich - 2/13/2008

Question
Dear Sir,
This has been the most painful course I have ever lived through. I really want to understand and I have tried very hard - I have been out of school way to long for this particular class - so any help on understanding would be much appreciated...I have already submitted this and failed miserably- I just want it to get you know? At your mercy - Anna Shepard

Determine whether the following equations have a solution or not? Justify your answer.
a) 5x2 + 8x + 7 = 0
b) (7)1/2y2 - 6y - 13(7)1/2 = 0
c) 2x2 + x - 1 = 0
d) 4/3x2 - 2x + 3/4 = 0
e) 2x2 + 5x + 5 = 0
f) p2 - 4p + 4 = 0
g) m2 + m + 1 = 0
h) 3z2 + z - 1 = 0


If x = 3 and x = -5, then form a quadratic equation.

What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.

Create a real-life situation that fits into the equation (x + 3)(x - 4) = 0 and express the situation as the same equation.


Answer
Questioner:   Anna
Category:  Advanced Math
Private:  No
 
Subject:  Quadratic equations and their applications
Question:  Dear Sir,
This has been the most painful course I have ever lived through. I really want to understand and I have tried very hard - I have been out of school way to long for this particular class - so any help on understanding would be much appreciated...I have already submitted this and failed miserably- I just want it to get you know? At your mercy - Anna Shepard

Determine whether the following equations have a solution or not? Justify your answer.
a) 5x2 + 8x + 7 = 0
b) (7)1/2y2 - 6y - 13(7)1/2 = 0
c) 2x2 + x - 1 = 0
d) 4/3x2 - 2x + 3/4 = 0
e) 2x2 + 5x + 5 = 0
f) p2 - 4p + 4 = 0
g) m2 + m + 1 = 0
h) 3z2 + z - 1 = 0

If x = 3 and x = -5, then form a quadratic equation.

What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.

Create a real-life situation that fits into the equation (x + 3)(x - 4) = 0 and express the situation as the same equation.
...................................
Hi, Anna,

Sorry this is so painful.  It sounds as if you need a lot of practice in certain things. Such as factoring.  A few hundred practice examples would be in order. (No, I'm not kidding -- I mean a few hundred.)  And before that, a few hundred examples of multiplying out binomials.

So, assuming you have done those (by the end of the week, I assume) here's what you will do:

1. Clear any fractions -- multiply by whatever is the LCD. (I hope you know what that means.)

2. Make the right side zero, if it isn't already.

3. Factor the left side. [After 200-300 examples, this will be a cinch.]
What if it isn't factorable?  
I know you learned the quadratic formula, (since you talked about the Discriminant) and it applies.

4. Set each factor SEPARATELY equal to zero and solve each.
That's it.


Determine whether the following equations have a solution SET or not? Justify your answer.

>> Well, if you put it that way, the answer is yes, because each is a quadratic.  But I think you wanted to FIND the solution set,

a) 5x2 + 8x + 7 = 0  

Since you talked about 'D', you might as well start using it.  Do it right, now:

A. Make sure the equation is in Standard FORM.  [BTW, don't confuse the word FORM with FORMULA.]

Form:  ax^2 + bx + c = 0

Formula:  D = b^2 - 4ac.

Application:  Match a,b,c, and WRITE DOWN WHAT THEY ARE.  

Substitute carefully.  USE PARENTHESES.

a = 5,  b = 8,  c = 7

D = 64 - 4(5)(7)  << use the parentheses EXACTLY as I just did. No more, no less.

D = 64 - 140

D = - 76    << Don't blow a sign here -- get it right.

Now use some rules to do some detective work:

If  D < 0 [that means 'is negative']
  then the roots are imaginary, and you can't factor.
If  D = 0,
  then the roots are real and identical, and you can factor.
If  D > 0 [that means 'is positive']
   then the roots are real, and
   IF D is a perfect square, [look that up]
       then the roots are rational, and you can factor.
       otherwise you cannot.

OK, apply:  D = -76. Negative. Roots imaginary, no factoring.

Use the Q.Formula:

   - b +- sqrt(b^2 - 4ac)
x = ---------------------- << long fraction line.
             2a

I'll leave the rest to you.

.............................
b) (7)1/2y2 - 6y - 13(7)1/2 = 0   << not clearly written.

c) 2x2 + x - 1 = 0  Can be factored, because D = 9, a perfect square.

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

2x - 1 = 0  << first equation

x + 1 = 0   << second

etc.
.............................
d) 4/3x2 - 2x + 3/4 = 0

Clear fractions:  Multiply by 12, your LCD, EVERY TERM.  Do it right:

(12)4/3x2 - (12)2x + (12)3/4 = 0

16x^2 - 24x + 9 = 0

D = 576 - 576 = 0, Factors and you have a 'double' root.
..........................
e) 2x2 + 5x + 5 = 0

D = - 15, so you have imaginary roots and need the Q.F.

........................
f) p2 - 4p + 4 = 0

g) m2 + m + 1 = 0

h) 3z2 + z - 1 = 0
I think you can handle these, now.
...........................................
If x = 3 and x = -5, then form a quadratic equation.

Just work things backwards:


x = 3     x = -5
x - 3 = 0,  x + 5 = 0
(x - 3)(x + 5) = 0
and multiply out.
.............................


Create a real-life situation that fits into the equation (x + 3)(x - 4) = 0 and express the situation as the same equation.

>> Sorry, but I retired from real-life situations a long time ago.  You're on your own here.