Chapter 8 More About Equations

Contents

8.2 Problems Leading to Quadratic Equations

8.3 Solving Simultaneous Equations by Algebraic Method

8.4 Graphical Solutions of Simultaneous Equations

8.5 More about Graphical Methods in Solving Simultaneous Equations

8 More about Equations

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8.1 Equations Reducible to Quadratic Equations

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More About Equations8

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A. Fractional Equations


Example 8.1T

.10136

x

x Solve

06

73

1036

36

10)1(36

xx

xx

xx

33

2

03023

or

or

x

xx

Solution:

0)3)(23(

0673 2

xx

xx

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B. Equations with Power More Than 2


31

0301

0)3)(1(

0322

or

or

y

yy

yy

yy

1

312

x

x (rejected) or

Example 8.3T

.032 24 equation the of roots real the Find xx

Solution:

have weSince ,2xy

becomes equation the Put 032, 242 xxxy

There is no real number x whose square is negative.

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C. Equations with Surd Form


Example 8.4T

.032 xx Solve

13

0103

0)1)(3(

0322

or

or

y

yy

yy

yy

9

13

x

x (rejected) or

becomes equation the Put 032, xxxy

have weSince ,xy

Solution:

Squaring both sides of an equation will sometimes create a number that is not a root of the original equation.

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D. Indical Equations


Example 8.5T

.06222 Solve xx

ana

aa

n

mnnm

loglog

Since y = 2x, we have

Put y = 2x, the equation 22x – 2x – 6 = 0 becomes

23

0203

0)2)(3(

062

or

or

y

yy

yy

yy

3log2log

2232

x

xx (rejected) or

Solution:

)2(58.12log

3logplacesdecimaltocorrectx

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E. Logarithmic Equations


Example 8.6T

.6log)1log(log xx Solve

When x = –3, log x and log (x+1) are undefined, therefore x = –3 is rejected.

6log)1(log

6log)1log(log

xx

xx

0)3)(2(

06

6)1(2

xx

xx

xx

(rejected) or

or

32

0302

x

xx

Solution:

MNNM logloglog

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Strategy for Solving Word Problems

1. Read the problem carefully – understand the problem; know what is given and what is to be found. If appropriate, draw figures or diagrams and label both known and unknown parts.

2. Let one of the unknown quantities be represented by a variable, say x, and try to represent all other unknown quantities in terms of x.

3. Set up an equation.

4. Solve the equation.

5. Check and interpret all solutions in the context of the originalproblem – not just for the equation found in Step 3.

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Example 8.8T

Consider a rectangle with an area of 100 cm2. If its length is 3 cm longer than its breadth, find the length of the rectangle. ( Give the answer correct to 2 decimal places. )

Solution:

01003

100)3(2

xx

xx

Let the length of the rectangle be x cm, then the width is (x – 3) cm.

= 11.61 (correct to 2 decimal places)

The length of the rectangle is 11.61 cm.

(rejected) or 2

4093

2

4093

12

)100(14)3()3( 2

x

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To solve a pair of simultaneous linear equations in two unknowns such as

The key step is to substitute the linear equation into the quadratic

equation to eliminate one of the two unknowns.

To solve a pair of simultaneous equations in two unknowns in which one is in linear form and one is in quadratic form, for example,

8.3 Solving Simultaneous Equations by Algebraic Method

quadratic

linear 2

,

43

1

xxy

xy

，

12

32

yx

yx

One method of solving them is to substitute one linear equation into the other one in order to eliminate one of the two unknowns.

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A. Solving Simultaneous Equations by Graphical Method


Solutions of two simultaneous equations are the solutions that satisfy both equations.

When solving a pair of simultaneous equations in two unknowns in which one

is linear and one is quadratic, we can draw the graph of each equation in the

same Cartesian coordinate plane.

The point(s) of intersection of the two graphs will give the solution(s) of the

two equations. However, they are only approximate solutions.

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B. Number of Points of Intersection of a Parabola and a Line


To solve a pair of simultaneous equations in which one is linear and the other is quadratic (in the form y = ax2 + bx + c, where a ≠ 0) by graphical method, the graphs of the parabola and the straight line may:

Case 1 : intersect at two distinct points, indicating that there are two different

solutions; orCase 2 : touch each other at one point only, indicating that there is only one

solution; or

Case 3 : have no intersections, indicating that there are no real solutions.

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Without the actual drawing of the graphs, the number of points of intersectionof the two graphs can be determined algebraically by the following steps:

Step 1 : Use the method of substitution to eliminate one of the unknowns (either x or y) of the simultaneous equations. We can then obtain a quadratic equation in one unknown.

Step 2 : Evaluate the discriminant (Δ) of the quadratic equation obtained

in Step 1.

• If Δ > 0, then there are two points of intersection.

• If Δ = 0, then there is only one point of intersection.

• If Δ < 0, then there are no intersections.

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Example 8.19T

Without solving the simultaneous equations algebraically, find the numberof points of intersection of the parabola y = 2x2 and the straight line y = 3x + 5.

Δ> 0 corresponds to the quadratic equation 2x2 – 3x – 5having two unequal real roots.

Solution:

)2..(....................53)1......(....................2 2

xyxy

)3(....................0532

2532

2

xx

xx

049

)5(24)3()3( 2

of

Substituting (2) into (1),

There are two points of intersection.

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8.5 More about Graphical Methods in Solving Simultaneous Equations

When we are given a graph of quadratic function such as y = x2, we can use it to solve any quadratic equation graphically such as x2 – x – 2 = 0 by the following procedures:

Step 1 : Write the equation as x2 = x + 2.

Step 2 : Hence, we can write this quadratic equation as two simultaneousequations ( one linear and one quadratic ) in two unknowns x and y,namely y = x2 and y = x + 2.

Step 3 : Draw the graphs of the two simultaneous equations in the same

Cartesian coordinate plane. The x-coordinates of their pointsof intersection will give the solutions of the quadratic equation x2 – x – 2 = 0.

Chapter 8 More About Equations

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