Question

Write and solve a quadratic equation for the situation below. Choose the answer that has both an equation that correctly models the situation as well as the correct solution for the situation. An isosceles right triangle has sides that are x + 2 units long and a hypotenuse that is 8 units long. What is the length of the missing sides of the triangle?

Answer 1

2(x^2 + 4x - 28) = 0.

The length of the missing sides are:

4√2 units.

or 5.66 units ( to the nearest hundredth).

Explanation:

Applying the Pythagoras Theorem:

8^2 = (x + 2)^2 + (x + 2)^2

2(x^2 + 4x + 4) = 64

2x^2 + 8x + 8 - 64 = 0

2x^2 + 8x - 56 = 0

2(x^2 + 4x - 28) = 0 models the situation.

Solving:

x = [- 4 +/- √(4^2-4*1*-28)] / 2

= (-4 +/- √128) / 2

= (-4 + 8√2) / 2 , (-4 - 8√2) / 2 (we ignore this negative root).

= -2 + 4√2.

This is 3.66 to the nearest hundredth.

So the length of the 2 equal sides is 2 + (- 2 + 4√2) = 4√2.

or 5.66 to the nearest hundredth.