Quadratic equations (F.4)

1.
Let t and u be the tens digit and the units digits respectively.

t = u - 5 ...... (1)
(10t + u)(t + u) = 234
[10(u - 5) + u][(u - 5) + u] = 234
(11u - 50)(2u - 5) = 234
22u2 - 155u + 250 = 234
22u2 - 155u + 16 = 0
u = [155 √(1552 - 4 x 22 x 16)]/(2 x 22)
u = (155 √22617)/44

Since u is not an integer, thus the question has no solution.


2.
(2a - 1)x2 - 8x + 6 = 0

The equation has two equal real roots.
D = 0
(-8)2 - 4(2a - 1)(6) = 0
64 - 48a + 24 = 0
-48a = -88
a = 11/6


3.
(1 - p)x2 - 2(2 - p)x + (1 - p) = 0

The equation has no real roots.
D < 0
[-2(2 - p)]2 - 4(1 - p)2 < 0
(2 - p)2 - (1 - p)2 < 0
[(2 - p) + (1 - p)] [(2 - p) - (1 - p)] < 0
3 - 2p < 0
p > 3/2


4.
y = (k - 1)x2 + 3x - 4 has two different x - intercepts.
Hence, (k - 1)x2 + 3x - 4 has two different real roots.
D > 0
(3)2 - 4(k - 1)(-4) > 0
9 + 16k - 16 > 0
16k > 7
k > 7/16


5.
(a)
y = (2k + 1)x2 - 4(k - 2)x + 16 touches the x-axis at only one point
Hence, (2k + 1)x2 - 4(k - 2)x + 16 = 0 has only equal real roots.
D = 0
[-4(k - 2)]2 - 4(2k + 1)(16) = 0
(k - 2)2 - 4(2k + 1) = 0
k2 - 4k + 4 - 8k - 4 = 0
k2 - 12k = 0
k(k - 12) = 0
k = 0 or k = 12

(b)
When k = 12:
[2(12) + 1]x2 - 4[(12) - 2]x + 16 = 0
25x2 - 40x + 16 = 0
(5x - 4)2 = 0
x = 4/5 (double roots)

x - intercept = 4/5
The graph touches the x-axis at (4/5, 0).


6.
(a)
x[(2k - 1)x + 3](3k + 1) + 9 = 0
[(2k - 1)x2 + 3x](3k + 1) + 9 = 0
(2k - 1)(3k + 1) + 3(3k + 1) + 9 = 0

x[(2k - 1)x + 3](3k + 1) + 9 = 0 has two equal real roots
(2k - 1)(3k + 1) + 3(3k + 1) + 9 = 0 has two equal real roots.
D = 0
[3(3k + 1)]2 - 4(2k - 1)(3k + 1)(9) = 0
(3k + 1)2 - 4(2k - 1)(3k + 1) = 0
3k + 1 ≠ 0, hence (3k + 1) - 4(2k - 1) = 0
3k + 1 - 8k + 4 = 0
-5k = -5
k = 1

(b)
k = 1
x{[2(1) - 1]x + 3}[3(1) + 1] + 9 = 0
x(x + 3)(4) + 9 = 0
4x2 + 12x + 9 = 0
(2x + 3)2 = 0
x = -3/2 (double roots)
=