Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

Topic closed

**Confused****Guest**

[Sorry, confused, I accidently deleted your question when I was doing a little cleaning up , please post it again, and I will try to solve it for you - MathsIsFun]

**Confused****Guest**

(That's ok.^^; I'll post it again)

I need help with two problems...

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

(x+2)(3)(x+2)-(x-2)(3)(2)x-2)(x+2)(-7)

[3x^2+12x+4]-[6x-12]=-7x^2+28

3x^2+12x+4-6x+12=-7x^2+28

3x^2+7x^2+12x-6x+12+4-28=0

10x^2+6x-12=0 I'm stuck here... (The answer is X={2/5,-1} but I can never seem to get it.)

7/(x^2-5x)+3/(5-x)=4/x

(5-x)(x)(7)+(x)(x^2-5x)(3)x^2-5x)(5-x)(4)

35x-7x^2+3x^3-15x^2=20X^2-100x-4x^3+20x^2

3x^3+4x^3-20x^2-20x^2-15x^2-7x^2+100x+35x=0

7x^3-62x^2+35x=0 Here is where I'm stuck... (Someone told me the answer was X=27/7, but I have no idea how to get there)

~Thank You!

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

OK, I didn't check your work, I will just assume that you got this far:

10x²+6x-12=0

That is a quadratic equation, because it looks like:

**ax² + bx + c = 0**

The answer to a quadratic equation is done using this formula:

**x = (-b ± √(b² - 4ac) / 2a**

So:

x = (-6 ± √(6² - 4*10*(-12)) / 2* 10 = -6 ± √(36 - (-480))) / 20 = (-6 +/- √(516))/ 20

Unfortunately, the square root of 516 is not a neat number, so I have to use my calculator to work out the two answers.

x = (-6 ± 22.72 ) / 20 = (-6 + 22.72 ) / 20 and (-6 - 22.72 ) / 20 = 0.836 and -1.436

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

So, perhaps I better check how you got there.

Start with: (x+2)/(x-2)-2/(x+2)=-7/3

Eliminate those nasty fractions by multiplying both sides by (x+2)(x-2):

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

Simplify: (x+2) * (x+2) - (x-2) * 2 = (x+2)(x-2) * (-7/3)

Simplify More: (x² + 4x + 4) - (2x-4) = (x² - 4) * (-7/3)

Simplify More: x² + 4x + 4 - 2x + 4 = (x² - 4) * (-7/3)

Simplify More: x² + 2x + 8 = (x² - 4) * (-7/3)

Multiply by 3: 3(x² + 2x + 8) = -7(x^2 - 4)

Expand: 3x² + 6x + 24 = -7x^2 + 28

Move all to Left Hand Side: 3x² + 6x + 24 +7x² - 28 = 0

Simplify: 10x² + 6x - 4 = 0

AHA! Now try the quadratic equation solution ...

x = (-b ± √(b² - 4ac) / 2a

x = (-6 ± √(6² - 4*10*(-4)) / 2* 10 = -6 ± √(36 - (-160))) / 20 = (-6 ± √(196))/ 20

x = (-6 ± 14)/ 20 = -20/20 and 8/20 = -1 and 2/5

DONE

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

I don't have time right now to solve the second one, but it should be much the same.

7/(x^2-5x)+3/(5-x)=4/x

Notice that (5-x) and (x^2-5x) are very similar - just multiply (5-x) times -x and you get (x^2-5x) - that would be where I would start.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**Confused****Guest**

Thank you so much for your help! I never knew there was a formula I was suppose to follow... I made sure to put it in my algebra notebook.:)

However, when I tried to solve the second prob. again I came out with a binomial instead...

7/(x^2-5x)+3/(5-x)=4/x

(7)+(-x)(3)=(-x)(5-x)(4/x)

(7)+(-3x)=4(x^2-5x)

x(7+(-3x))=4(x^2-5x)

-3x^2+7x=4x^2-20x

-3x^2+7x-4x^2+20x=0

-7x^2+27x=0 Here is where I can't seem to expand the binomial...

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Don't worry, binomials are even easier than quadratics!

You had: -7x^2+27x=0

Factorise the x: x(-7x+27)=0

Solve: x=0 or -27/7.

You already used stuff that is much harder than this to get where you got to, so this shouldn't be any problem.

If you're still confused (which you shouldn't be) then you can still use the quadratic equation, taking c=0.

Why did the vector cross the road?

It wanted to be normal.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

Hmmm ... you may not have been shown the "quadratic formula" because you were meant to rearrange the problem till a solution is obvious, like mathsyperson did for the second problem. I am not very good at that, and always aim for the quadratic style of solution.

Offline

**Confused****Guest**

Oh, thanks mathsyperson! Your right binomials do seem easier! Thanks to both of you for all your help!:D

Pages: **1**

Topic closed