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.

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...

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.

]]>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

]]>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

]]>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!

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