algebra chapter 1 lch gh a roche. simplify (i) = = = multiply each part by x factorise the top...
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Algebra
Chapter 1LCH GH
A Roche
Simplify (i)
=
=
=
multiply each part by x
factorise the top
Divide top & bottom by (x-3)
p.3
+Simplify (ii)
+ Multiply second part above and below by -1So that both denominators are the same
Factorise the top
p.3
p.3Simplify 1. 4x(3x2 + 5x + 6) – 2(10x2 + 12x)
= 12x3 + 20x2 + 24x – 20x2 - 24x= 12x3
4x(3x2 + 5x + 6) – 2(10x2 + 12x)
p.3Simplify 2. (x + 2)2 + (x - 2)2 - 8
= (x2 + 4x + 4) + (x2 - 4x + 4) - 8= 2x2
(x + 2)2 + (x - 2)2 - 8 Expand the squares
p.3Simplify 3. (a + b)2 - (a - b)2 – 4ab
= (a2 + 2ab + b2) - (a2 – 2ab + b2) – 4ab= 0
Expand the squares (a + b)2 - (a - b)2 – 4ab= a2 + 2ab + b2 - a2 + 2ab - b2 – 4ab
p.3Simplify 4. (2a + b)2 – 4a(a + b)
= (4a2 + 4ab + b2) - 4a2 – 4ab= b2
Expand (2a + b)2 – 4a(a + b)= 4a2 + 4ab + b2 - 4a2 – 4ab
p.3Factorise5. x2 + 3x= x(x + 3) x2 + 3x HCF
Factorise6. 3xy – 6y2
3xy – 6y2 HCF= 3y(x - 2y)
p.3Factorise7. a2b + ab2
= ab(a + b) a2b + ab2 HCF
Factorise8. 9x2 – 16y2
9x2 – 16y2 Difference of 2 squares
= (3x – 4y)(3x + 4y)
p.3Factorise9. 121p2 – q2
= (11p – q)(11p + q) 121p2 – q2 Difference of 2 squares
Factorise10. 1 – 25a2
1 – 25a2 Difference of 2 squares
= (1 – 5a)(1 + 5a)
p.3Factorise11. x2 – 2x - 8
= (x )(x ) x2 – 2x - 8 Quadratic factors
Factorise12. 3x2 + 13x - 103x2 + 13x - 10 Quadratic
Factors= (3x – 2)(x + 5)Check!+15x
-2x
-8(1)(-8)(2)(-4)(4)(-2)(8)(-1)
-10(1)(-10)(2)(-5)(5)(-2)
(10)(-1)
Which factors add to -2?= (x +2 )(x - 4 )
= (3x )(x )
p.3Factorise13. 6x2 - 11x + 36x2 - 11x + 3 Quadratic
Factors= (3x – 1)(2x - 3)Check!-9x
-2x+3
(1)(3)(-1)(-3)
= ( )( ) 6x2
(6x)(x)(3x)(2x)
= (3x )(2x )
p.7 Example
(i) If a(x + b)2 + c = 2x2 + 12x + 23, for all x, find the value of a, of b and of c.
a(x + b)2 + c = 2x2 + 12x + 23
a(x2 + 2xb + b2) + c = RHS
Expand the LHS
ax2 + 2axb + ab2 + c = RHSObserve that the LHS is a Quadratic Expression in x
(a)x2 + (2ab)x + (ab2 + c) = 2x2 + 12x + 23
Equate coefficients of like terms
a = 2 2ab = 12
2(2)b = 12
4b = 12
b = 3
ab2 +c = 23
(2)(3)2 +c = 23
18 +c = 23
c = 5
p.7 Example
(ii) If (ax + k)(x2 – px +1) = ax3 + bx + c, for all x, show that c2 = a(a – b).
(ax + k)(x2 – px + 1) = ax3 + bx + c
ax(x2 –px + 1) +k(x2 –px + 1) = RHS
Expand the LHS
ax3 - apx2 + ax + kx2 –kpx + k = RHS
(a)x3 + (-ap + k)x2 + (a - kp)x + k = ax3 + 0x2 + bx + c
Equate coefficients of like terms
a = a k - ap = 0 a – kp = b k = c
c - ap = 0
c = ap
p = c/a
a – c(c/a) = b
a – c2 /a = b
a – b = c2 /a
a(a – b) = c2
p.9 Example
Write out each of the following in the form ab, where b is prime:(i) 32 (ii) 45 (iii) 75
Divide by the largest square number:
149
162536496481
100121144169
(i) 32
(iii) 75
(ii) 45
= (16 x 2) = 162 = 42
= (9 x 5) = 35
=(25 x 3) = 53
p.9 Example
Express in the form , a, b N :
(iv) (v)
Divide by the largest square number:
149
162536496481
100121144169
(iv)
(v)
p.9 Example
(vi) Express in the form k2.
(vi)
p.9 Example
(i) Express 18 + 50 - 8 in the form ab, where b is prime.
(ii) 20 - 5 + 45 = k5; find the value of k.Divide by the largest
square number:149
162536496481
100121144169
(i) 18 + 50 - 8
(ii) 20 - 5 + 45
= (9 x 2) + (25 x 2) - (4 x 2)
= 62
= 25 - 5 + 35
= 45
= 32 + 52 – 22
= (4 x 5) - 5 + (9 x 5)
Examples of Compound Surds
P.10
a- b
a- b
a + b 1 + 5
3- 24
13- 7
Conjugate Surds
P.10
a - b
a + b
a + b
Compound Surd Conjugate 1 Conjugate 2
a - b - a + b
a + b
a - b - a + b
-a - b
When a compound surd is multiplied by its conjugate the result is a rational number.
We use this ‘trick’ to solve fractions with compound denominators
p.10 Example
Show that
Multiply top and bottom by conjugate of denominator
Note that the bottom is difference of 2 squares
Q.E.D.
1 – 3
p.11
Solving Simultaneous Equations
For complicated simultaneous equations we use the substitution-elimination method
p.12 ExampleSolve for x and y the simultaneous equations:
x + 1 – y + 3 = 4, x + y – 3 = 1 2 3 2 2
Get rid of fractions by Multiplying
x + 1 – y + 3 = 4 2 3
(6)(x + 1)– (6)(y + 3) = (6)4 2 3
x + y – 3 = 12 2
(3)(x + 1)– (2)(y + 3) = 24
3x + 3 – 2y - 6 = 24
3x– 2y = 27
2x + 2(y – 3) = 2(1) 2 2
2x + y – 3 = 1
2x + y = 4
Now solve these simultaneous equations in the normal way
x = 5 and y = -6
p.12-13 Example
Solve for x, y and z: x + 2y + z = 35x – 3y +2z = 193x + 2y – 3z = -5
Label the equations 1, 2 & 3
2
1
3
21
Eliminate z from 2 equations
-2x - 4y -2z = -65x – 3y +2z = 19
x -2
3x - 7y = 13 4
3x + 2y – 3z = -53x + 6y + 3z = 9
3
1 x 3
6x + 8y = 4 5
Now solve simultaneous equations 4 & 5 in the usual way
Sub these values into equation 1
x + 2y + z = 3
(2) + 2(-1) + z = 3
z = 3
x = 2
y = -1
We find that:
p.13
Note:
If one equation contains only 2 variables then the other 2 equations are used to obtain a second equation with the same two variables
Here, from equation 1 and 2, y should be eliminated to obtain an equation in x and z, which should then be used with equation 3
3x + 2y - z = -35x – 3y +2z = 35x + 3z = 14
e.g. Solve
3
2
1