F.4-5 Maths

1.
Let a m and b m be the lengths of sides of the two squares.
(a > b)

4a + 4b - 2b = 16 …… [1]
a² + b² = 13 …… [2]

From [1]:
4a + 2b = 16
2a + b = 8
b = 8 - 2a …… [3]

Put [3] into [2]:
a² + (8 - 2a)² = 13
a² + 64 - 32a + 4a² = 13
5a² - 32a + 51 = 0
(a - 3)(5a - 17) = 0
a = 3 or a = 17/5
a = 3 or a = 3.4

Put a = 3 into [3] :
b = 8 - 2(3)
b = 2

Put a = 3.4 into [3] :
b = 8 - 2(3.4)
b = 1.2

The lengths of the sides of the two square is:
2 m and 3 m or 1.2 m and 3.4 m


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2.
xy = 45 …… [1]
(x + 2)(y -1) = 45 - 1 …… [2]

From [1]:
x = 45/y …… [3]

Put [3] into [2]:
[(45/y) + 2](y - 1) = 44
45 - (45/y) + 2y - 2 = 44
2y - 1 - (45/y) = 0
y[2y - 1 - (45/y)] = 0
2y² - y - 45 = 0
(y - 5)(2y + 9) = 0
y = 5 or y = -9/2 (rejected)

Put y = 5 into [3]:
x = 45/5
x = 9

Hence, x = 9, y = 5


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3.
L: x + y + k = 0 …… [1]
C: y = 2x² - 4x + k …… [2]

From [1]:
y = -x - k …… [3]

[2] = [3]:
2x² - 4x + k = -x - k
2x² - 3x + 2k = 0

Since the equation has two real roots, thus determinant Δ > 0
(3)² - 4(2)(2k) > 0
9 - 16k > 0
-16k > -9
16k < 9
k < 9/16


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4.
(a)
C: y = px² + 2x + 6 …… [1]
L: y = 8 - 2x …… [2]

[1] = [2]:
px² + 2x + 6 = 8 - 2x
px² + 4x - 2 = 0 …… [3]

Since the equation has double roots, thus determinant Δ = 0
(4)² - 4(p)(-2) = 0
16 + 8p = 0
p = -2


(b)
Put p = -2 into [3]:
-2x² + 4x - 2 = 0
x² - 2x + 1 = 0
(x - 1) = 0
x = 1 (double roots)

Put x = 1 into [2]:
y = 8 - 2(1)
y = 6

Coordinates of A = (1, 6)


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5.
3x - 2y - 1 = 0 …… [1]
5x² - y² = 1 …… [2]

From [1]:
2y = 3x - 1
y = (3x - 1)/2 …… [3]

Put [3] into[2]:
5x² - [(3x - 1)/2]² = 1
5x² - (3x - 1)²/4 = 1
20x² - 9x² + 6x - 1 = 4
11x² + 6x - 5 = 0
(11x - 5)(x + 1) = 0
x = 5/11 or x = -1

Put x = 5/11 into [3]:
y = [3(5/11) - 1]/2
y = 2/11

Put x = -1 into [3]:
y = [3(-1) - 1]/2
y = -2

x = 5/11, y = 2/11 or x = -1, y = -2