Posted by **astrid** on Wednesday, April 2, 2008 at 10:08pm.

A ladder is resting against a wall. The top of the ladder touches the wall at a height of 15 feet and the length of the ladder is one foot more than twice the distance from the wall. Find the distance from the wall to the bottom of the ladder. (Hint: Use the Pythagorean Theorem.)

I guess the wall and the ladder form a right triangle.

- algebra -
**Reiny**, Wednesday, April 2, 2008 at 10:15pm
yes, so solve

15^2 + x^2 = (2x+1)^2

- algebra -
**astrid**, Wednesday, April 2, 2008 at 10:31pm
I tried that but it didn't give me a quadratic equation and that's what we've been studying.

- algebra -
**Reiny**, Wednesday, April 2, 2008 at 10:40pm
why did you not get a quadratic?

15^2 + x^2 = (2x+1)^2

225 + x^2 = 4x^2 + 4x + 1

3x^2 + 4x - 224 = 0

see?

since you are studying quadratics I will let you finish it. (hint: 8 is a nice number)

- algebra -
**barb**, Wednesday, April 2, 2008 at 10:48pm
2(3y)2(2x)

- algebra -
**Anonymous**, Wednesday, April 2, 2008 at 10:46pm
? Just move all values to one side and leave the other side to zero and there's your quadratic formula.

Let x be the distance from the wall to the bottom of the ladder

Let 2x+1 be the length of the ladder

15^2 + x^2 = (2x+1)^2

4x^2 + 4x + 1 = 15^2 + x^2 I just expanded and switched the sides

3x^2 + 4x - 224 = 0

x^2 + 4x - 672 = 0 <-- bridge method is not in the NYC curriculum but I don't know about yours

(x+28)(x-24) = 0

(3x+28)(3x-24) = 0

(3x+28)(x-8) = 0

x = 8, -28/3(rej because its silly for the distance to be negative)

- algebra -
**astrid**, Wednesday, April 2, 2008 at 10:53pm
I get it now. I messed up on (2x+1)^2. I didn't simplify it right. Thank you guys so much.

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