Theory of Equations Test

Result

Theory of Equations Test - 2

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Question 1
1 / -0

The number of ordered pairs of natural numbers (x, y) satisfying the equation 2x + 3y = 100 is

A
13

B
14

C
15

D
16

Solution

2x + 3y = 100
The first pair, (x, y), satisfying the above equation is (2, 32).
The other ordered pairs can be derived from the above pair by following a pattern (Adding 3 to every value of x and subtracting 2 from every value of y).
Thus, the pairs are (5, 30); (8, 28); (11, 26); (14, 24); (17, 22); (20, 20); (23, 18); (26, 16); (29, 14); (32, 12); (35, 10); (38, 8); (41, 6); (44, 4); (47, 2). The total pairs are 16.
Question 2
1 / -0

Examine the nature of roots of
(i) x² + 9x + 27 = 0
(ii) 6x² - 13x - 5 = 0

A
(i) Real (ii) Real

B
(i) Real (ii) Imaginary

C
(i) Imaginary (ii) Real

D
(i) Imaginary (ii) Imaginary

Solution

(i) D = b2 - 4ac = (+9)2 - 4 ´ 1 ´ 27 = 81 - 108 = - 27, which is
negative. Hence, the roots will be imaginary and conjugate.
(ii) D = b2 - 4ac = (- 13)2 - 4 ´ 6 ´ (- 5) = 169 + 120 = 289,
which is a perfect square. Hence, the roots will be real, rational
and unequal.
Question 3
1 / -0

If both x and y are real, and x² + y² + 2x – 10y = –26, then find the value of x + y.

A
1

B
2

C
3

D
4

Solution

x2 + y2 + 2x – 10y + 26 = 0
⇒ (x + 1)2 + (y – 5)2 = 0
Since both x and y are real, so (x + 1)2 = 0.
⇒ x = –1
And (y – 5)2 = 0 ⇒ y = 5

∴ x + y = 4