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 November 21st, 2014, 10:34 AM #1 Newbie   Joined: Nov 2014 From: Cottingham Posts: 8 Thanks: 0 Math Focus: Surds! Fractions! With powers! Or equations with powers and a number I have knowledge of solving fractions without powers, e.g. (2x+2)/3+x/5=10. But for (2x(x+1))/2=20, I got to 2x^2+2x=40, what do I do next?
November 21st, 2014, 11:51 AM   #2
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Quote:
 Originally Posted by alifeee I got to 2x^2+2x=40, what do I do next?
That's a classical second degree equation.

\displaystyle \begin{aligned} & 2x^2 + 2x -40 = 0 \\ & x^2 + x - 20 = 0 \Rightarrow ax^2 + bx + c \\ & x = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ \end{aligned}

You can continue..

 November 21st, 2014, 12:24 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond $\displaystyle x^2+x-20=(x+5)(x-4)$
November 21st, 2014, 03:47 PM   #4
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Quote:
 Originally Posted by alifeee But for (2x(x+1))/2=20, I got to 2x^2+2x=40, what do I do next?
$\displaystyle \color{blue}{\dfrac{\not{2}x(x+1)}{\not{2}}=20\ \Longleftrightarrow\ x(x+1)=20 \\\;\\ \\\;\\ x\ \ and\ \ x+1\ \ can\ be,\ here,\ two\ consecutive\ integers. \\\;\\ 4\cdot5 = 20 \\\;\\ (-5)\cdot(-4)=20}$

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