Hey, I'm having trouble with a few of these.

1. x^2 - 6x + 3 = 0

2. x^2 + 5x + 1 = 0

3. 2x^2 - 50 = 0

4. x = 3/x + 7

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- Sep 22nd 2006, 11:17 AMcarlito1Equations Help
Hey, I'm having trouble with a few of these.

1. x^2 - 6x + 3 = 0

2. x^2 + 5x + 1 = 0

3. 2x^2 - 50 = 0

4. x = 3/x + 7 - Sep 22nd 2006, 11:28 AMtopsquark
Is the problem which method to use or using the methods? In general, if you can't figure out how to solve a quadratic, there's always the quadratic formula to fall back on. (This is going to look terrible without LaTeX, sorry!)

ax^2 + bx + c = 0 implies

x = -b/(2a) (+/-) [sqrt(b^2 - 4ac)]/(2a) (Where the "+/-" is "plus or minus.")

The method of "completing the square" always works as well. Then there's the methods you can use to factor, but at a glance I don't think any of these factor.

I'll do this one by completing the square, if you need help with the quadratic formula, just let me know. :)

x^2 - 6x + 3 = 0

x^2 - 6x = -3

We want to add a number to both sides such that the LHS is a perfect square. The format for this is the identity:

(x + a)^2 = x^2 + (2a)x + a^2.

So, looking at the form (with the "a" in it) and your problem the coefficient of the x term is - 6 = 2a. Thus a = -3. So we want to add (-3)^2 to both sides of the equation:

x^2 - 6x + 9 = -3 + 9

(x - 3)^2 = 6

x - 3 = (+/-)sqrt(6)

x = 3 (+/-) sqrt(6).

-Dan - Sep 22nd 2006, 11:33 AMtopsquark
We can do this one using completing the square as well.

x^2 + 5x + 1 = 0

x^2 + 5x = -1

The coefficient of x is 5 = 2a, so a = 5/2. Thus we need to add (5/2)^2 to both sides:

x^2 + 5x + (5/2)^2 = -1 + (5/2)^2

(x + 5/2)^2 = -1 + 25/4 = -4/4 + 25/4 = 21/4

x + 5/2 = (+/-)sqrt(21/4) = (+/-)sqrt(21)/2

x = -5/2 (+/-) sqrt(21)/2.

I'll do this with the quadratic formula as well:

x^2 + 5x + 1 = 0

Here a = 1, b = 5, c = 1.

x = -b/(2a) (+/-) [sqrt(b^2 - 4ac)]/(2a) = -5/(2*1) (+/-) [sqrt(5^2 - 4*1*1)]/(2*1)

x = -5/2 (+/-) [sqrt(25 - 4)]/2 = -5/2 (+/-) sqrt(21)/2.

-Dan - Sep 22nd 2006, 11:38 AMtopsquark
Mostly I'd like you to try these again based on the work in the last two posts, but I'll get you started.

3. 2x^2 - 50 = 0

We really don't need the completing the squares technique here. I lied in my first post, this one factors. You would start the same way in either case:

2x^2 - 50 = 0

x^2 - 25 = 0

This is the difference between two perfect squares. You should have a formula to factor this in your book. (Or complete the square, but in this case a = 0, which is a little silly. However moving the constant term to the other side of the equation gives: x^2 = 25. You don't need to fiddle with the LHS because it's already a perfect square! So just take the square root. Remember the "+/-"!)

4. x = 3/x + 7

The only "hidden" difficulty here is that we can't have a solution where x = 0. (We don't in any case, but it's always good to check this.)

I'm going to multiply both sides of this equation by x:

x^2 = 3 + 7x

x^2 - 7x - 3 = 0.

This one doesn't factor, but can you take it from here?

-Dan - Sep 24th 2006, 05:06 AMcarlito1
I get x=9.25 or x= 4.71 for number 4 - can someone check this

- Sep 24th 2006, 05:30 AMOReilly
- Sep 24th 2006, 07:45 AMQuick
- Sep 24th 2006, 10:39 AMtopsquark
Multiply both sides by x:

x^2 = 3 + 7x

x^2 - 7x = 3

To complete the square, 2a = -7 so a = -7/2. Thus we add a^2 = 49/4 to both sides.

x^2 - 7x + (-7/2)^2 = 3 + (-7/2)^2

(x - 7/2)^2 = 3 + 49/4 = 12/4 + 49/4 = 61/4

x - 7/2 = (+/-)sqrt(61/4) = (+/-)[sqrt(61)]/2

x = 7/2 (+/-) [sqrt(61)]/2

Numerically this is x = 7.405124838 or x = -0.405124838

-Dan