Algebra 1

I have this problem. In my book it doesnt have any examples of this type of problem. Please help!

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Radical Equations
    _________    _______
  \/ x^2 + 2x = \/ 10 - x

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Square both sides of the equations and then you've got a quadratic equation.

x^2 + 2x = 10 - x
x^2 + 3x - 10 = 0
(x+5)(x-2) = 0
x = -5 or 2

-_-;

Uh... can you explain?

How did 2x turn into 3x?

(x+5)(x-2) = 0 ? where did that come from?

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How did 2x turn into 3x?

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He "moved" it from the right side to the left side. Look at this equation:

3 = 2 + 1

That is the same as:

3 - 1 = 2

You can add, subtract, multiply and divide with whatever you want only if you do the same on both sides. What Jason did was add x on both sides.

2x + x turned into 3x
and -x + x turned into 0 (nothing)

so -2 and 5 are factors of 10. Which is number c.

the sum of the factors of c have to equal b wich is 3.

-2 + 5 = 3..

Did I get it?

okay... now lets say I have a problem like this?

Code:

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  ______
\/ x + 6 = x + 4

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that has a radical expresion on just one side... can I work it like this?

x + 6 = x^2 + 16

:confused:

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(x+5)(x-2) = 0 ? where did that come from?

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x^2 + 3x - 10 = 0
Original equation

x^2 + (2 * (3 / 2)) x - 10 = 0
2 * (3 / 2) is the same as 3. Multiply by 2 and then divide with 2 and you get the same number as before :)

x^2 + (2 * (3 / 2)) x + (3 / 2)^2 - (3 / 2)^2 - 10 = 0
First you add (3 / 2)^2, then you subtract it, so you haven't changed the value of the equation

(x + (3 / 2))^2 - (3 / 2)^2 - 10 = 0
Using the rule (a + b)^2 = a^2 + 2 * a * b + b^2 (sorry, don't know the english name of it :() you can change the first three terms of step 3 into the blue coloured part in this step

(x + (3 / 2))^2 - (9 / 4) - (40 / 4) = 0
3^2 equals 9 and 2^2 equals 4. I also change 10 into (40 / 4), which is the same thing

(x + (3 / 2))^2 - (49 / 4) = 0
-9 - 40 equals -49

(x +(3 / 2))^2 = (49 / 4)
Add (49 / 4) to both sides (see my previous post)

x + (3 / 2) = +/- sqrt( 49 / 4 )
Take the square root of both sides. Notice the +/-, since both a and -a is a^2 when squared (is this a real word?).

x = -(3 / 2) +/- (7 / 2)
Subtract (3 / 2) from both sides

x1 = -(10 / 2) = -5
x2 = (4 / 2) = 2
Since we have a +/-, x have two values (unless it's a double root). -3 - 7 is -10, and -3 + 7 is 4

If you put in any of these values in (x + 5)(x - 2) = 0, you see that the left side has the value 0, and the equation is correct.

Hope this helps a little. Algebra can be tough sometimes ;).

:eek: ............................................

(Vicious is suffering from a braon freeae that Magos has caused him)

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Originally posted by Vicious
can I work it like this?

x + 6 = x^2 + 16

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No, the correct would be:

x + 6 = (x + 4)^2

x + 6 = x^2 + 2 * 4 * x + 4^2

I wasn't sure how much explaining to do since I wasn't sure at what level you were in in Algebra I.

--edited--

Once you get x^2 + 3x - 10 = 0 you can use the equation

Code:

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                    _________
x = -b (+/-) \/ b^2 - 4ac
      --------------------------
                2a

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Where a is what is in front of x^2 (1), b is what is in front of x (3) and c is -10

Thanks for your help..

I guess I might can get through with Algebra before next school year starts :(

Try #math on DALnet if you need help with math :)

if you like this wait until calculus.
f(x) = x^3+2x^2+5x+3
synthetic division, first and second derivatives, and all that fun stuff :)