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Solutions to the exercises, specified in the example of the ExSol package

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Solutions to the exercises, specified in the example of the ExSol package

Walter Daems October 23, 2018

Exercise 2-1: Solve the following equation for x ∈ C, with C the set of complex numbers:

5x

2

− 3x = 5 (1)

Solution: Let’s start by rearranging the equation, a bit:

5.7x

2

− 3.1x = 5.3 (2)

5.7x

2

− 3.1x − 5.3 = 0 (3)

The equation is now in the standard form:

ax

2

+ bx + c = 0 (4)

For quadratic equations in the standard form, we know that two solutions exist:

x

1,2

= −b ± √ d

2a (5)

with

d = b

2

− 4ac (6)

If we apply this to our case, we obtain:

d = (−3.1)

2

− 4 · 5.7 · (−5.3) = 130.45 (7) and

x

1

= 3.1 + √ 130.45

11.4 = 1.27 (8)

x

2

= 3.1 − √ 130.45

11.4 = −0.73 (9)

The proposed values x = x

1

, x

2

are solutions to the given equation.

Exercise 2-2: Consider a 2-dimensional vector space equipped with a Euclidean distance func- tion. Given a right-angled triangle, with the sides A and B adjacent to the right angle having lengths, 3 and 4, calculate the length of the hypotenuse, labeled C.

Solution: This calls for application of Pythagoras’ theorem, which tells us:

kAk

2

+ kBk

2

= kCk

2

(10)

and therefore:

kCk = q

kAk

2

+ kBk

2

(11)

= p

3

2

+ 4

2

(12)

= √

25 = 5 (13)

Therefore, the length of the hypotenuse equals 5.

1

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