Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Question and Answer

Questions

Symbol Defintion

Sqr(x) = Square root of x

Q01 | - Equation theory : quartic equation

Q02 | -

Q03 | -

Q04 | -

Q05 | -

Q06 | -

Q07 | -

Q08 | -

Q09 | -

Q10 | -

Q11 | - Quiz

Q12 | - Quiz answer

Q13 | - Reference

Answers

Q01. Equation theory : Quartic equation

Question

Equation : x^4 + b*x^3 + c*x^2 + d*x + e = 0
Roots are 1, 2, 3, 4. Find d

Keywords

Let roots be p,q,r,s
1. Coefficients of x is p*q*r + p*q*s + p*r*s + q*r*s = -d/a
2. Coefficients of x is (1/p + 1/q + 1/r + 1/s) = -d/e

Method 1 : Use formula 2

1/1 + 1/2 + 1/3 +1/4 = -d/e and e = p*q*r*s = 1*2*3*4 = 24
Hence d = -24*(1 + 1/2 + 1/3 + 1/4) = -(24 + 12 + 8 + 6) = -50

Mehtod 2 : Use formula 2 with a = 1

d/a = -(1*2*3 + 1*2*4 + 1*3*4 + 2*3*4) = -(6 + 8 + 12 + 24) = -50

Method 3 : x^4 + b*x^3 + c*x^2 + d*x + e = (x-1)*(x-2)*(x-3)*(x-4)

(x-1)*(x-2)*(x-3)*(x-4)
= (x^2 - 3*x + 2)*(x^2 - 7*x + 12)
= x^4 - 10*x^3 + 35*x^2 - 50*x + 24
Hence x = -50

Go to Begin

Q02. x^4 - 10*x^3 + 35*x^2 - 50*x + 24 = 0 has roots 1 and 2, find other roots
Method 1 : Use equation theory

Let other roots be p and q
Hence 1 + 2 + p + q = -(-10)/1 or p + q = 7 .... (1)
Hence 1*2*p*q = 24/1 or p*q = 12 ............... (2)
Solve equation (1) and (2), we have p = 3 and q = 4

Method 2 : Use synthetic division

1 -10 +35 -50 +24 | 1
... 1 -09 +26 -24
------------------
1 -09 +26 -24 +00 | 2
... 2 -14 +24
-------------
1 -07 +12 +00
Hence (x-1)*(x-2)*(x^2 - 7*x + 12) = 0
Hence x^2 - 7*x + 12 = 0
Hence (x - 3)*(x - 4) = 0
Hence x = 3 or x = 4

Q03. Answer

Q04. Answer

Q05. Answer

Q06. Answer

Q07. Answer

Q08. Answer

Q09. Answer

Q10. Answer

Q11. Quiz for Equations

Questions

1. Solve Abs(x-2) = 3.
2. Solve x^2 - 3*x + 2 = 0.
3. Solve x^3 - 6*x^2 + 13*x - 12 = 0.
4. Solve x^4 - 10*x^3 + 37*x^2 - 64*x + 48 = 0.
5. Solve x^4 - 16 = 0.
6. p,q,r,s are roots of x^4 - 10*x^3 + 37*x^2 - 64*x + 48 = 0. Find p*q*r*s.
7. Solve x^4 + x^3 + x^2 + x + 1 = 0 graphically.
8. Solve e^x + 2*e^(-x) - 1 = 0.
9. Find the value of ln(e^2.5) - 2.5.
10 How many real roots in Abs(x^2 - 6*Abs(x) + 8) = 1/2.

Go to Begin

Q12. Answers of quiz for Equations

Answers

1. Solve Abs(x - 2) = 3.

x - 2 = +3 and x = +5.
x - 2 = -3 and x = -1.

2. Solve x^2 - 3*x + 2 = 0.

(x - 1)*(x - 2) = 0 and x= 1 or 2

3. Solve x^3 - 6*x^2 + 13*x - 12 = 0.

Use synthetic division. Trial values are 1, -1, 2, -2, 3, -3, 4, -4, ....

4. Solve x^4 - 10*x^3 + 37*x^2 - 64*x + 48 = 0.

Use synthetic division. Rrial values 1, -1, 2, -2, .....

5. Solve x^4 - 16 = 0.

16 = 2^4 and use factors of x^4 - a^4

6. p,q,r,s are roots of x^4 - 10*x^3 + 37*x^2 - 64*x + 48 = 0. Find p*q*r*s.

Product of roots = 48/1.

7. Solve x^4 + x^3 + x^2 + x + 1 = 0 graphically.

Draw a large unit circle (one unit = 10 cm).
Draw angles 72, 144, 216 and 288 and cut circle at A, B, C, D.
Measure the coordiantes of A, B, C, D
Hence x1 = a1 + i*b1, ....

8. Solve e^x + 2*e^(-x) - 1 = 0.

(e^x)^2 - e^x + 2 = 0. Solve this equation using factors.

9. Find the value of ln(e^2.5) - 2.5.

Use ln(e^x) = x.

10 How many real roots in Abs(x^2 - 6*Abs(x) + 8) = 1/2.
- Study subject Graph of y = Abs(x^2 - 6*Abs(x) + 8)

Go to Begin

AL 11 13. Reference

1. Solve polynomial equation on PC computer : MD2002 Section 17.
2. Equations including absolute on internet : See keywords absolute.

Go to Begin

Show Room of MD2002

Contact Dr. Shih

Math Examples Room