# (Wolframalpha) this shouldnt give x=50?

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:

x-100*y/2=2500

x-100*y=5000

y=5100/x


and then:

5100/x=100+2x

5100=100+2x*x

5000=2x^2

sqrt(2500)=x

50=x


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]


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

how can I get the same results on wolframalhpa ?

thank you ! (:

• Wolfram's answer is correct based on your input. Could you maybe show your steps? – icurays1 Nov 20 '12 at 18:10
• 5100/x=100+2x 5100=100+2x*x 5000=2x^2 sqrt(2500)=x 50=x :) – Alvarolm Nov 20 '12 at 18:11
• Ah, there should be an $x$ attached to the 100 as well: $5100=100x+2x^2$. Now use the quadratic formula. – icurays1 Nov 20 '12 at 18:13
• why is it wrong ? – Alvarolm Nov 20 '12 at 18:20
• 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 $5100=(100+2x)x=100x+2x^2$. – 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 $y = 100$. 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:

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