English test-1S5 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
English test-1S5 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
English test-1S5 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
English test-1S6 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: Draw the graph of: y = 4x² − 4x – 2. Explain each step using an accurate vocabulary.
English test-1S6 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: Draw the graph of: y = 4x² − 4x – 2. Explain each step using an accurate vocabulary.
English test-1S6 Exercise 1: Solve each equation by completing the square: 1. 𝑥² + 15𝑥 + 12 = 0 2. 4𝑥² + 5𝑥 = −1 Exercise 2: We watch a video about a problem posed by the Babylonians. What was it? Describe the method the Babylonians stated in the Ancient Times. Explain their reasoning. Exercise 3: Draw the graph of: y = 4x² − 4x – 2. Explain each step using an accurate vocabulary.
Correction:
Exercise 1 : 1. 𝑥 2 + 15𝑥 + 12 = 0 First, we move the constant term to the right-hand-side of the equation 𝑥² + 15𝑥 = −12 We then have to complete the square on the left-hand-side of the equation. For that, we need a constant term that turns the expression into a perfect square trinomial. Since the middle term in a perfect square trinomial is always twice the product of the square root of the other two terms, we re-write our expression as: 𝑥² + 2 ×
15 2
15
× 𝑥 = −12
We see that the constant we are seeking must be ( 2 ) ² 15
By adding ( 2 ) ² to both sides of the equation, this can be factored as: (𝑥 + (𝑥 +
15 2 ) 2 15 2 2
= −12 +
) =
225 4
177 4
we just have to make the calculation on the right, and we get:
Then we have to take the square root of both sides:
𝑥+
15 2
=√
177 4
or
𝑥+
15 2
𝑥=
−15+√177 2
or
𝑥=
−15−√177 2
= −√
177 4
−15+√177 −15−√177 ; } 2 2
We can conclude by writing the set of solutions between braces: 𝑆 = {
2. 4𝑥² + 5𝑥 = −1 To solve this equation, we first have to divide all terms by the coefficient of the 𝑥² term: 5
1
5
1
𝑥² + 4 𝑥 = − 4
We then use the same process as question 1. Rewrite the expression:
𝑥² + 2 × 8 𝑥 = − 4
In order to have a perfect square trinomial on the left-hand-side we 5 2 8
need to add the constant ( ) to both sides of the equation. Then we factor and simplify: 5 2
1
25
9
(𝑥 + 8) = − 4 + 64 = 64 Take the square root of both sides: 5
3
3 8
5 8
𝑥+8=8 𝑥= − =−
1 4
5
3
or
𝑥 + 8 = −8
or
𝑥 = − − = −1 So the set of solution is: 𝑆 = {− ; −1}
3 8
5 8
Exercise 2: In the video we watched, Marcus du Sautoy was presenting a problem posed by the Babylonians: If a field has an area of 55 units, and one side is 6 units longer than the other, how long is the shorter side?
1 4
The Babylonian solution was to reconfigure the field as a square. Cut 3 units of the end and move this round.
Now, there is a 3 by 3 piece missing, so let’s add this in. The area of the field has increased by 9 units. This makes the new area 64. So the sides of the square are 8 units.
The problem solver knows he added 3 to the side, so the original width must be 5. That way, the length has to be 11. Exercise 3: Since we know that the triangle is a right-angled triangle, we can use the Pythagorean Theorem. As 5 is the hypotenuse, we know that: 𝑥² + (𝑥 + 1)2 = 5² We just have to expand the left-handside of the equation. 𝑥² + 𝑥² + 2𝑥 + 1 = 25 Then, we simplify and move the constant term on the right hand side. 2𝑥² + 2𝑥 = 24
We divide all terms by 2
𝑥² + 𝑥 = 12
and we rewrite the x term , and add 4 to both sides
1 2
1
1 4
𝑥² + 2 × × 𝑥 + = 12 +
1 4
we can now factor the perfect square trinomial into a perfect
square binomial 1 2
(𝑥 + 2) = 𝑥=
7−1 2
49 4
=3
By taking the square root of both sides of the equation, we get: or 𝑥 =
−7−1 2
= −4
As 𝑥 represents a length, it can’t be negative, so x must be equal to 3. The area of the triangle is therefore,
3×4 2
= 6 squared-units.
Exercise 3: y = 4x² − 4x – 2 is the equation of a parabola written in standard form. To make it easier to graph, I am going to convert it into the vertex form. I first move the constant term to the left-hand side of the equation, and then, by completing the square, I will write the right hand side as a perfect binomial square.
𝑦 + 2 = 4(𝑥 2 − 𝑥) I rewrite the x term 1
1
𝑦 + 2 = 4 (𝑥 − 2 × 2 × 𝑥) to complete the square, I need to add 4 in the brackets, but since the bracket is factored by 4, it’s the same as adding 1. So I have to add it to the other side as well. 1 2
𝑦 + 3 = 4 (𝑥 − 2)
And this is the vertex form. 1
Therefore, I know that the vertex is at coordinates: (2 ; −3) and so I can plot this first point on the graph. Then I need to find the x-intercept. There can be none, 1 or 2 solutions. 1 2
I set y to 0 : 3 = 4 (𝑥 − 2)
and I just have to solve this equation by dividing by 4 and then
taking the square root of both sides. 1
𝑥−2= 𝑥=
√3 2
√3+1 2
≈ 1.4
1
and
𝑥−2=−
and
𝑥=
−√3+1 2
√3 2
≈ −0.4
We can plot those 2 points (I labelled them B
and C) We have to look for the y-intercept, there is always one and only one. We have to substitute x for 0 in the equation. It’s much more appropriate to use the standard form: y = 4x² − 4x – 2 is equivalent to 𝑦 = −2 when 𝑥 = 0 We can add this point (D) to the graph. Since a parabola is mirror-image, I can find another point using the y-intercept. Since we moved 1 unit to the left from the vertex to plot point D, we have to move one unit to the right, keeping the same height to plot point E. To conclude, we just have to join the 5 points in a smooth curve.
