Question

A ladder is resting againist a building. The distance from the bottom of the ladder is 18ft less then the length. How high up is the ladder resting againist the wall if the distance is 1ft less then the length of the ladder?

Answer 1

let L= length ladder
distance from wall = L-18
height up the wall = L - 1

by pythaporas
L^2 = ( L-18)^2 + ( L-1)^2

solving for L and adding 1 will give you the answer

Answer 2

sorry - its subtract 1

Answer 3

I have gotton as far as (x-18)^2+(x-1)^2=x^2
x^2-36x+324+x^2-2x+2=x^2
x^2-38x+326=0
After that I am soo lost!

Answer 4

ok - theres a slight error on second line its +1 = x^2

giving x^2 - 38x + 325 = 0

Answer 5

325 = 5 * 5 * 13
so the correct factors kook like 13 and 25 - they add up to 38
so its (x-25)(x-13) = 0

x = 25 or 13
cant be 13 because of the x-18

length of ladder = 25
and of height of ladder on wall = 25-1 = 24 ft

Answer 6

How did you get to the (x-25)(x-13)=0

Answer 7

ok x^2 - 38x + 325 = 0
to factor we need two numbers which when multiplied give 325 and when added give -38

with big number like 325 its best to find its prime factors
325 = 5 * 5*13
- first divide by 5 giving 65 and 65 = 5*13
now 5*5 = 25
we need -38 so its -25 and -13
giving (x-25)(x-13) = 0

Answer 8

you ok with that drgn?