This is my problem:

I need x or y for the triangle area that forms between the vertical axis(y) and the function y=100+2x where the area is equal to 2500.

so I used for condition to the linear function:

knowing that the triangle area in this case should be like: x*y/2=area, so:




and then:






the weird thing is that works for any area, and gives me the correct result for what I'm looking for, wich is x=50 and y=f(50)=200, if the area is calculated as is shown in the condition: 200-100*50/2=2500 !

5100/x=100+2x [http://www.wolframalpha.com/input/?i=5100%2Fx%3D100%2B2x][1]

it outputs x=-5 (5+sqrt(127)) and x=5 (-5+sqrt(127))

how can I get the same results on wolframalhpa ?

thank you ! (:

  • $\begingroup$ Wolfram's answer is correct based on your input. Could you maybe show your steps? $\endgroup$ – icurays1 Nov 20 '12 at 18:10
  • $\begingroup$ 5100/x=100+2x 5100=100+2x*x 5000=2x^2 sqrt(2500)=x 50=x :) $\endgroup$ – Alvarolm Nov 20 '12 at 18:11
  • 3
    $\begingroup$ Ah, there should be an $x$ attached to the 100 as well: $5100=100x+2x^2$. Now use the quadratic formula. $\endgroup$ – icurays1 Nov 20 '12 at 18:13
  • $\begingroup$ why is it wrong ? $\endgroup$ – Alvarolm Nov 20 '12 at 18:20
  • 1
    $\begingroup$ When you multiply both sides of an equation by something, you have to distribute. Put parentheses around it and you'll see: $(5100/x)=(100+2x)$ so $5100=(100+2x)x=100x+2x^2$. $\endgroup$ – icurays1 Nov 20 '12 at 18:23

You made an error when multiplying $x$ on both sides.

You had $$\frac{5100}{x} = 100 + 2x$$

In order to remove the $x$ from the denominator you correctly decided to multiply by $x$ on both sides. However, when you do this you should get $$x \frac{5100}{x} = x(100 + 2x)$$ And multiplying $x$ through the equation becomes $$ 5100 = 100x + 2x^2$$ and now I'm sure you can solve it!

EDIT: To address the edit in your question, be careful when setting up the area of this triangle because it is actually $$\frac{1}{2} x(y-100)$$ since the bottom part of our triangle is located at $y = 100$. our triangle!

And to solve this area to be 2500, we would plug in $$ \frac{1}{2} x(y-100) = 2500$$ And we know that $y = 100 + 2x$, so plugging that in for $y$ gives us $$ \frac{1}{2}x(100 + 2x - 100) = 2500$$ and canceling out the $100$s and multiplying through by the $x$ and $\frac{1}{2}$ gives $$ x^2 = 2500$$ or $x = 50$.


You start with the equation $\frac{5100}{x} = 100 + 2x.$ If the left side and the right side are equal, then I can do the same to both sides and they'll still be equal. Let's multiply both sides by $x$. I get $5100 = 100x + 2x^2.$ Bringing all of the terms over to one side, we get $2x^2 + 100x - 5100 = 0.$ Next, notice that there is a common factor: $2(x^2 + 50x - 2550) = 0.$ Finally, we use the quadratic formula where $a = 1,$ $b = 50$ and $c = -2550$. We have:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \, ,$$ $$x = \frac{-50 \pm \sqrt{50^2-4\times 1 \times (-2550)}}{2 \times 1} \, , $$ $$x = \frac{-50 \pm \sqrt{12700}}{2} \, , $$ $$x = \frac{-50 \pm 10\sqrt{127}}{2} \, , $$ $$x = -25 \pm 5\sqrt{127} \, .$$

It seems that the website was correct. Notice that $-5(5\pm \sqrt{127}) = -25 \mp 5\sqrt{127}.$ If you don't see why $\sqrt{12700} = 10\sqrt{127}$ then notice that $12700 = 2^2 \times 5^2 \times 127$, where $127$ is prime and so:

$$\sqrt{12700} = \sqrt{2^2 \times 5^2 \times 127} = \sqrt{2^2} \times \sqrt{5^2} \times \sqrt{127} = 2 \times 5 \times \sqrt{127} = 10\sqrt{127} \, . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.