Algebra, inequalities and counting

Written by Hassan Raza (Mathematics)

NET · Mathematics · 9 questions

Name: ____________________ Date: ____________

What is the sum of the squares of the roots of x² - 3x + 2 = 0?
- A 5
- B 3
- C 4
- D 9
If x² - 5x + 6 = 0 has roots alpha and beta, what is alpha² + beta²?
- A 25
- B 11
- C 5
- D 13
How many terms are there in the expansion of (a + b)⁸?
- A 7
- B 9
- C 10
- D 8
The sum of the first n natural numbers is 210. What is n?
- A 15
- B 21
- C 19
- D 20
Which values of x satisfy |x - 2| < 3?
- A x > -1
- B x < 5
- C x < -1 or x > 5
- D -1 < x < 5
What is the remainder when x³ - 2x² + x - 5 is divided by x - 2?
- A 0
- B 3
- C -5
- D -3
What is log base 10 of 0.001?
- A 3
- B -2
- C -3
- D 0.001
In how many different ways can the letters of the word BOOK be arranged?
- A 12
- B 8
- C 24
- D 6
What is the value of 5C2?
- A 25
- B 20
- C 5
- D 10

Answer key

1. A, 2. D, 3. B, 4. D, 5. D, 6. D, 7. C, 8. A, 9. D

The roots are 1 and 2, so the sum of the squares is 1 + 4 = 5.
Sum = 5 and product = 6, so alpha² + beta² = 25 - 12 = 13.
The expansion of (a + b)^n has n + 1 terms, so here there are 9.
n(n + 1) / 2 = 210 gives n(n + 1) = 420, which is true for n = 20.
|x - 2| < 3 means -3 < x - 2 < 3, so -1 < x < 5.
By the remainder theorem, substitute x = 2: 8 - 8 + 2 - 5 = -3.
0.001 = 10⁻³, so its logarithm is -3.
There are 4 letters with the letter O repeated twice, so the number of arrangements is 4! / 2! = 12.
5C2 = 5! / (2! x 3!) = 10.