Word problem for Quadratic Equations

Question

A ladder of length 3x+2feet is positioned against a wall such that the bottom is x−2feet away from a wall. The distance between the floor and the top of the ladder is 3xfeet.



Find the length, in feet, of the ladder. Assume that a right angle is formed by the wall and the floor.

Answer 1

Okay, Tisha, you and I can manage this one together. Ready?

You have a right triangle, so the Pythagorean Theorem applies:

A² + B² = C²

If A is the base of the triangle, we have A = X-2

That makes B the height of the ladder on the wall: B = 3X

Finally, our ladder is the hypothesis: C = 3X + 2

Substitute them into the formula:

(X - 2)² + (3X)² = (3X + 2)²

X² - 4X + 4 + 9X² = 9X² + 12X + 4

10X² -4X +4 = 9X² + 12X + 4

We have a quadratic equation, so it must equal zero. Gather everything on one side:

-9X²-12X-4    -9X² -12X - 4

 X² -16X  = 0

Now we factor and solve:

X (X -16) = 0

X = 0  ,  X = 16

Since the triangle base would be zero when X is zero, that answer is extraneous. Only X = 16 works, so we substitute it into each function:

A = 16 - 2 = 14

B = 3(16) = 48

C = 3(16) +2 = 48 + 2 = 50

All done!

Answer 2

Use the pythagorean theorem which says that c^2=a^2+b^2 where c is the hypotenuse (longest side).  So

(3x+2)^2=(3x)^2+(x-2)^2

9x^2+12x+4=9x^2+x^2-4x+4

12x+4x-x^2=0

x^2-16x=0

x(x-16)=0

x=0 or x=16

If x= 0 then the ladder bottom would have distance (0) - 2 = -2 which makes no sense and so x is 16 feet. Substitute x back into the expression 3x+2 and we get 3(16) + 2 = 50 feet. Therefore the ladder is 50 feet long.