急! F5 About equations 1exq07

47.
One of the roots of the equation of 2x² + ax - 12b = 0 is -3.
Put x = -3 into the equation 2x² + ax - 12b = 0
2(-3)² + a(-3) - 12b = 0
18 - 3a - 12b = 0
3a + 12b = 18 ...... [1]

The equation x² - ax + b = 0 has two equalreal roots. Δ = 0
(-a)² - 4(1)(b) = 0
a² - 4b = 0
12b = 3a² ...... [2]

Put [2] into [1] :
3a + 3a² = 18
a² + a - 6 = 0
(a + 3)(a - 2) = 0
a = -3 or a = 2

When a = -3 :
Put a = -3 into [1] :
3(-3) + 12b = 18
b = 9/4

When a = 2 :
Put a = 2 into [1] :
3(2) + 12b = 18
b = 1

Hence, a = -3 and b = 9/4
or a = 2 and b = 1


48.
(a)
According to the graph, (0, 18) lies on the parabola.
Put x = 0 and y = 18 into the equation y = ax² + bx + c :
18 = a(0) + b(0) + c
c = 18

(b)
The equation is : y = ax² + bx + 18

y = ax² + bx + 18 touches y = 0(x-axis).
Hence, ax² + bx + 18 = 0 has only oneroot. Δ = 0
b² - 4a(18) = 0
b² - 72a = 0
b² = 72a

(c)
(3, 0) lies on the parabola :
Put x = 3 and y = 0 into y = ax² + bx + 18 :
(0) = a(3)² + b(3) + 18
9a + 3b + 18 = 0 ...... [1]

From the reaction of part (b) :
b² = 72a
a = b²/72 ...... [2]

Put [2] into [1] :
9(b²/72) + 3b + 18 = 0
(b²/8) + 3b + 18 = 0
b² + 24b + 144 = 0
(b + 12)² = 0
b = -12 (double roots)

Put b = -12 into [2] :
a = (-12)²/72
a = 2

Hence, a = 2 and b = -12


49.
Area of ΔOAB is (12 and 1/4) :
(1/2)ab = (12*4 + 1)/4
ab = 49/2 ...... [1]

(2, 3) lies on AB.
Slope of AB :
(3 - b)/(2 - 0) = (3 - 0)/(2 - a)
(3 - b)/2 = 3/(2 - a)
(3 - b)(2 - a) = 6
6 - 3a - 2b + ab = 6
ab - 3a - 2b = 0 ...... [2]

Put [1] into [2] :
(49/2) - 3a - 2b = 0
6a + 4b = 49
4b = 49 - 6a
b = (49 - 6a)/4 ...... [3]

Put [3] into [1] :
a(49 - 6a)/4 = 49/2
49a - 6a² = 98
6a² - 49a + 98 = 0
(2a - 7)(3a - 14) = 0
a = 7/2 or a = 14/3

Put a = 7/2 into [1] :
(7/2)b = 49/2
b = 7

Put a = 14/3 into [1] :
(14/3)b = 49/2
b = 21/4

When a = 7/2 and b = 7, the equation of AB is :
[x/(7/2)] + y/(7) = 1
2x + y - 7 = 0

When a = 14/3 and b = 21/4, the equation of AB is :
[x/(14/3)] + [y/(21/4)] = 1
9x + 8y - 42 = 0