Quadratic equations

Question

One leg of a right triangle is 7 cm less than the length of the other leg. The length of the hypotenuse is 17 cm. Find the lengths of the legs.

Answer 1

x and (x - 7) are legs. Then by Pythagorian Theorem: x² + (x - 7)² = 17²; 2x² - 14x - 240 = 0; x² - 7x - 120 = 0; (x - 15)(x + 8) = 0; x = 15 cm, x - 7 = 8 cm

Answer 2

The hypotenuse is 17 so let's call the missing sides a and b.

From the Pythagorean Theorem we have

a² + b² = 17²

We know one side is 7 less than the other side so let's just say a = b - 7 (we could easily use b = a - 7, it doesn't matter).

a = b - 7

Substituting in for a

(b - 7)² + b² = 17²

Expanding

b² - 14b + 49 + b² = 289

2b² - 14b + 49 - 289 = 0

2b² - 14b - 240 = 0

Divide by 2

b² - 7b -120 = 0

So we need to check to see if we can easily find factors of -120 that sum up to -7

Factors of 120:

1 and 120, 2 and 60, 3 and 40, 4 and 30, 5 and 24, 6 and 20, 8 and 15, 10 and 12.

Of those, there is one that when combined by addition or subtraction equals -7 and that's 8 and 15. 15 - 8 = 7, or more appropriately, -15 + 8 = -7.

So we have our factors.

(b - 15)(b + 8) = 0

So b = 15 or b = -8. Since we can't have a negative length, b = 15 is our answer.

Checking

The length of one side is 7 less than the other. 15 - 7 = 8 check.

a² + b² = 17²

8² + 15² = 17²

64 + 225 = 289

289 = 289 check!