Solving for x?

Question

x^2 - x - 6 = 0


x^2 - 29x + 100 = 0


5x^2 +100 = 0


x^2 + 36 = 0


5x^2 + 7x + 3


x^2 + 3x + 9 = 0


i have no idea where to start with these..need to know the HOW TO

Answer 1

well maybe it would be a good idea to go to class, pay attention in class, and do the homework. eventually, cheating wll catch up with you. The longer you try to get away with it, the bigger the price you'll pay.

Answer 2

When you solve for X or anyother variable for that matter you need to get X sitting by itself on one side of the equation with everything else on the other side. The X is kind of anti-social in that regard. It just wants to be alone.

There are several methods that you can use to help the X achieve its goal of being lonely. Looks to me like most of the exercises that you have here are use a technique called factoring. Polynomials are just big fancy expressions for numbers. If you know your times tables you can break 'normal' numbers like 9 or 24 into their component parts. So 9 = 3 * 3 = 9 * 1 and 24 = 12 * 2 = 24 * 1 = 6 * 4 = 3 * 2 * 4 = 8 * 3 = 2 * 2 * 2 * 3. Similarly you can factor a polynomial.

Generally the way to factor a polynomial is to start by looking at the last term which is usually a constant (or a number like 100, not a variable like x. We can't do much with x). After breaking it down into it's different factors like we did above (so 100 = 10*10, 20*5, 50*2, 25*4) we look at the middle term, usually something times x. What we want is a pair of factors that adds or subtracts to equal the middle term. Sounds confusing but it's not. Here's an example.

Take the second problem x^2 - 29x + 100 = 0.

We've already figured out our factors for 100, those being 10*10, 20*5, 50*2, 25*4. Our middle term is 29. 10+10 = 20 so that's not the factors we want, 20+5 = 25 so that's not it, 50+2 = 52 so that's right out and we're left with 25+4 = 29. Yay!

Uh oh! You say, it's a -29, not a plus 29. Not a big deal. Remember that -25*(-4) still equals 100 and -25+(-4) = -29 so we're safe with our choices. Now to put it all together.

The factored solution to our equation is:

(x-25)(x-4) = 0

If you're familiar with the FOIL (First Outer Inner Last) technique for expanding a factor polynomial then you can think of this as the reverse of that. In fact we can check our work by foiling this all out.

(x-25)(x-4) = x*x - 25x - 4x + (-25)(--4) = x^2 - 29x + 100 = 0.

So now the question, now that I"ve done all this factoring junk how do I solve for x? Have I just been wasting my time?

Well, look at our equation: (x-25)(x-4) = 0

When is this equal to 0? The answer is when either one of my terms containing x = 0. So either (x-25) = 0 or when (x-4) = 0 cuz after all 0*anything = 0. So then we can solve both of those for x and we'll have our final answer. At this point it's easy enough to solve for x as we merely need to add something to both sides of the equation:

(x-25) = 0 is first.

I add 25 to both sides to get

(x - 25) + 25 = 0 + 25. The 25s on the left side cancel to get me:

x = 25.

Similarly, for (x-4) = 0, x = 4.

This gives me my solution. The equation x^2 - 29x + 100 = 0 when x = 4 and when x = 25, or when x = {4, 25}.

Hope this helps.

Answer 3

You need to get x alone on one side of the equation when there is only one x in the problem. in the third problem:

5x^2 + 100 = 0
5x^2 = -100
x^2 = -100/5
but then you have to take the square root but you can;t take the square root of a negative number.

x^2 + 36 = 0
x^2 = -36
but again you can't take the square root of a negative number unless you said the answer was x = 6i where i =sqrt(-#)

for all the other problems you need to apply the quadratic formula or factor it out.

quadratic formula is x = -b +or- sqrt(b^2 - 4ac/(2a) where ax^2 + bx + c = 0

Answer 4

Here's a hint. Since you cant get a negative number as the result of a squre, you use IMAGINARY NUMBERS. So: i^2= minus 1. now if you look at

x^2 +36 = 0

you can solve it in this manner:

x^2 = -36

x=6i

Now use this idea for the rest.

Good luck!

Answer 5

Factoring 1. (x-3)(x+2) = 0 , so x=3 and x = -2 will be our solutions

Quadratic equation 29 +/- sqrt(29^2 - 4*1*100)
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''''''''''''''''''''''''''''''''''''' 2

3. No Real solution +/-sqrt(20) * i

4. No Real solution +/- 6*i

5. No real solution

6. No real solution

Answer 6

try www.webmath.com it gives you the answer, but tells you how to do it, first!!!

Answer 7

there is no other way