16sccmm6

1. CLASSICAL ALGEBRA

2. REMOVAL OF TERMS
• SOLVE BY REMOVING II TERM 𝑿 𝟒
− 𝟖𝑿 𝟑
− 𝑿 𝟐
+ 𝟔𝟖𝑿 + 𝟔𝟎 = 𝟎
• S0LUTION ;
• Replace x by (x+h)
𝒙 + 𝒉 𝟒
− 𝟖 𝒙 + 𝒉 𝟑
− 𝒙 + 𝒉 𝟐
+ 𝟔𝟖 𝒙 + 𝒉 + 𝟔𝟎 = 𝟎
𝒙 𝟒
+ 𝒉 𝟒
+ 𝟔𝒙 𝟐
𝒉 𝟐
+ 𝟒𝒙 𝟑
𝒉 + 𝟒𝒙𝒉 𝟑
− 𝟖𝒙 𝟑
− 𝟖𝒉 𝟑
− 𝟐𝟒𝒙 𝟐
𝒉 − 𝟐𝟒𝒙𝒉 𝟐
− 𝒙 𝟐
− 𝒉 𝟐
− 𝟐𝒙𝒉 + 𝟔𝟖𝒙 + 𝟔𝟖𝒉 + 𝟔𝟎 = 𝟎
𝒙 𝟒
+ 𝒙 𝟑
𝟒𝒉 − 𝟖 + 𝒙 𝟐
𝟔𝒉 𝟐
− 𝟐𝟒𝒉 − 𝟏 + 𝒙 𝟒𝒉 𝟑
− 𝟐𝟒𝒉 𝟐
− 𝟐𝒉 + 𝟔𝟖 + 𝒉 𝟒
− 𝟖𝒉 𝟑
− 𝒉 𝟐
+ 𝟔𝟖𝒉 + 𝟔𝟎 =0

3. 4h-8=0
h=2
1 -8 -1 68 60
0 2 -12 -26 84
2
1441 -6 -13 42
0 2 -8 -42
01 -4 -21
0 2 -4
--251 -2
0 2
1
0
Hence remove the second term increase the roots of the equation by 2

4. REMOVAL OF TERMS
𝑋4
− 25𝑋2
+ 144 = 0
Let x = 𝑥2
𝑥2
− 25𝑥 + 144 = 0
(x-16) (x-9)=0
X=16 , 9
X= ±4, ±3
X=x+h
X=4+2,-4+2, 3+2, -3+2
The roots of this given equation X= 6, -2, 5, -1.

5. 2)SOLVE BY REMOVING II TERM 𝒙 𝟑 − 𝟐𝟏𝒙 𝟐 + 𝟏𝟒𝟒𝒙 − 𝟑𝟐𝟎 = 𝟎
• Solution:
• Let replace x by (x+h)
𝒙 + 𝒉 𝟑 − 𝟐𝟏 𝒙 + 𝒉 𝟐 + 𝟏𝟒𝟒 𝒙 + 𝒉 − 𝟑𝟐𝟎 = 𝟎
𝒙 𝟑
+ 𝒉 𝟑
+ 𝟑𝒙𝒉 𝟐
+ 𝟑𝒙 𝟐
𝒉 − 𝟐𝟏𝒙 𝟐
− 𝟐𝟏𝒉 𝟐
− 𝟒𝟐𝒙𝒉 + 𝟏𝟒𝟒𝒙 + 𝟏𝟒𝟒𝒙𝒉 − 𝟑𝟐𝟎 = 𝟎
𝒙 𝟑 + 𝒙 𝟐 𝟑𝒉 − 𝟐𝟏 + 𝒙 𝟑𝒉 𝟐 + 𝟒𝟐𝒉 + 𝟏𝟒𝟒 − 𝟐𝟏𝒉 𝟐 + 𝟏𝟒𝟒𝒉 − 𝟑𝟐𝟎 = 𝟎
3h-21=0
h = 7

6. remove the second term increase the roots of the equation by 7
7 1 -21 144 -320
0 7 -98 322
1 -14 46 2
0 7 -49
1 -7 -3
0 7
1 0
𝑋3
+ 0𝑋2
− 3𝑋 + 2 = 0

7. (𝑥 − 1)(𝑥2 + 𝑥 − 2) = 0
X=1, 1, -2
1 0 -3 2
0 1 1 -2
0
1 1 -2
1
x = x+h
X=1+7, -2+7, 1+7
The roots of this given equation
8, 8, 5

8. 3)SOLVE BY REMOVING II TERM𝒙 𝟑 − 𝟏𝟐𝒙 𝟐 + 𝟒𝟖𝒙 − 𝟕𝟐 = 𝟎
• Solution :
• Replace x by (x+h)
𝒙 + 𝒉 𝟑 − 𝟏𝟐 𝒙 + 𝒉 𝟐 + 𝟒𝟖 𝒙 + 𝒉 − 𝟕𝟐 = 𝟎
𝒙 𝟑
+ 𝒉 𝟑
+ 𝟑𝒙 𝟐
𝒉 + 𝟑𝒙𝒉 𝟐
− 𝟏𝟐𝒙 𝟐
− 𝟏𝟐𝒉 𝟐
− 𝟐𝟒𝒙𝒉 + 𝟒𝟖𝒙 + 𝟒𝟖𝒉 − 𝟕𝟐 = 𝟎
𝒙 𝟑
+ 𝒙 𝟐
𝟑𝒉 − 𝟐 + 𝒙 𝟑𝒉 𝟐
− 𝟐𝟒𝒉 + 𝟒𝟖 + 𝒉 𝟑
− 𝟏𝟐𝒉 𝟐
+ 𝟒𝟖𝒉 − 𝟕𝟐 = 𝟎
3h-12=0
h = 4
Remove the second term increase the roots of the equation by 4

9. 1 -12 48 -72
0 4 -32 64
4
-81 -8 16
0 4 -16
01 -4
0 4
0
𝑥3
− 8 = 0
𝑥3
= 8 𝒙 = 𝟐

10. 1 0 0 -8
0 2 4 8
01 2 4
2
𝑥2
+ 2𝑥 + 4 = 0
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
a=1 b=2 c= 4
𝑥 =
−2 ± 4 − 4(4)
2(1)

11. 𝑥 =
−2 ± −12
2
𝑥 =
−2 ± 2 −3
2
𝑥 =
−1 ± −3
1
x = (x+h)
x = 2+4,
−1± −3
1
+4
Hence the roots of this equation 6, 3±√-3