Darkness

A quadratic with roots 1 and -1 could be written as

(x-1)(x+1) = 0

Expand the brackets and we get:

x^2 +0x -1 = 0 

So, the solution is 0 and -1

a - b = 1

a = b + 1

a^3 = (b+1)^3

= b^3 +3b^2 +3b + 1

As mentioned in the question above, a^3 - b^3 = 1

We have proven a^3 = b^3 +3b^2 +3b + 1

So, replacing a^3 with b^3 +3b^2 +3b + 1:

b^3 +3b^2 +3b + 1 - b^3 = 1

Collecting like terms

3b^2 + 3b = 0

3b^2 = -3b

b^2 = -b

b = -1, b = 0

If b = -1:

a-(-1) = 1

a + 1 = 1

a = 0

If b = 0:

a - 0 = 1

a = 1

So if b = -1, a = 0

If b = 0, a = 1

Part (a)

It can be either -1 * 0

Which equals 0

Or 0 * 1

Which equals 0

So the only answer is 0

Part (b)

It can be either -1 + 0

Which equals -1

Or 0 + 1

Which equals 1

So the answers are 1, -1

Part (c)

The possible values for a are 0, 1

The possible values for b are -1, 0

Another solution for this problem is using algebra.

Erwus's solution was probably cleaner, but I might as well post mine.

Let n be the side length of the square

n^2 = 9

(kn)^2 = 25

k^2*n^2 = 25

k^2 * 9 = 25

k^2 = 25/9

k = √25/9

= 5/3 or - 5/3

So 5/3 +(-5/3) = 0

800mL acid + 200mL water = 1000mL of total liquid.

Draining 150mL of total liquid

150/1000 = 15%, which means 15% of the water and 15% of the acid will be drained.

0.85* 800 = 680mL acid

0.85* 200 = 170mL water

= 680mL acid and 170mL water

Adding 150mL water

170 + 150 = 320mL water

So there would be 680mL acid and 320mL water.

Draining 250mL of total liquid

680 + 320 = 1000mL total liquid

250/1000 = 25%, which means 25% of the water and 25% of the acid will be drained.

0.75*680 = 510mL acid

0.75*320 = 240mL water

Adding 250mL water

240 + 250 = 490mL water

So your final answer is 490mL water.

Let's say a = 25% of the season

Let's say n is also 25% of the season.

a=35% win rate

3a/3 = (35*3)/3 % win rate

= 35%

(3a+n)/4 = 50%

3a+n = 200%

If a = 35%, 3a = 105%

105% + n = 200%

n = 95%

So, they must win 95% of the remainder of their games to finish the season with the same number of wins as losses.

I haven't explained this very well, but I hope you understand.

Nevermind. Nothing here.

a - b = 1

a = b + 1

a^3 = (b+1)^3

= b^3 +3b^2 +3b + 1

As mentioned in the question above, a^3 - b^3 = 1

We have proven a^3 = b^3 +3b^2 +3b + 1

So, replacing a^3 with b^3 +3b^2 +3b + 1:

b^3 +3b^2 +3b + 1 - b^3 = 1

Collecting like terms

3b^2 + 3b = 0

3b^2 = -3b

b^2 = -b

b = -1, b = 0

If b = -1:

a-(-1) = 1

a + 1 = 1

a = 0

If b = 0:

a - 0 = 1

a = 1

So if b = -1, a = 0

If b = 0, a = 1

Part (a)

It can be either -1 * 0

Which equals 0

Or 0 * 1

Which equals 0

So the only answer is 0

Part (b)

It can be either -1 + 0

Which equals -1

Or 0 + 1

Which equals 1

So the answers are 1, -1

Part (c)

The possible values for a are 0, 1

The possible values for b are -1, 0

x^2 - 8x + 15 + x^2 - 4x + 3

Collect like terms

= 2x^2 - 12x + 18 

= 2x^2 - 6x + 18 - 6x

= 2x(x-3) - 6(x-3)

= (2x-6)(x-3)

2x = 6

x=3

or x-3 = 0

x=3

So the only solution is x = 3

x^2 - 5x + 4 = 0

x^2 - x + 4 - 4x = 0

x(x-1) - 4(x-1) = 0

(x-4)(x-1)=0

x=4, x=1

9(x+3) = -8 + x

Expand brackets

9x + 27 = -8 + x

Collect like terms

8x = -35

x = -35/8

2x/(x+2) = x/(x-4)

Get rid of the divisors

2x(x-4) = x(x+2)

Expand

2x^2 - 8x = x^2 +2x

Collecting like terms

x^2 - 10x = 0

Factoring

x(x-10) = 0

x=0, x=10