0
$\begingroup$

I came across this problem on math stack exchange and tried to solve it myself: The position of a ladder leaning against a wall and touching a box under it.

What I did was set up 2 equations

$(x+1)^2+(y+1)^2=4^2$

$x^2+1^2=(4-(1+y)^2)^2$

Then by similar triangles:

$x/(x+1)=1/(1+y)$

I took $(x+1)^2+(y+1)^2=4^2$ then solved for 1+y to get $sqrt(-x^2-2x-15)$

I then plugged this back into the original and got

$x+1/x = 1/(-x^2-2x-15)$

However, my equation gives complex solutions. So where have I gone wrong? How can I get to the right solution using these 2 equations?

$\endgroup$
  • $\begingroup$ Please choose a more descriptive title. Thanks. $\endgroup$ – user370967 Nov 4 '18 at 8:37
1
$\begingroup$

The problem is in the step in which you solve the equation $(x+1)^2+(y+1)^2=4^2$. You should have obtained$$y+1=\pm\sqrt{15-x^2-2x}.$$

| cite | improve this answer | |
$\endgroup$

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.