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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 Home 8.1 Equations Reducible to Quadratic Equations

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More About Equations 8 Home Content P. 2 A. Fractional Equations 8.1 Equations Reducible to Quadratic Equations Example 8.1T Solution:

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More About Equations 8 Home Content P. 3 B. Equations with Power More Than 2 8.1 Equations Reducible to Quadratic Equations Example 8.3T Solution: There is no real number x whose square is negative.

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More About Equations 8 Home Content P. 4 C. Equations with Surd Form 8.1 Equations Reducible to Quadratic Equations Example 8.4T 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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More About Equations 8 Home Content P. 5 D. Indical Equations 8.1 Equations Reducible to Quadratic Equations Example 8.5T Since y = 2 x, we have Put y = 2 x, the equation 2 2x – 2 x – 6 = 0 becomes Solution:

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More About Equations 8 Home Content P. 6 E. Logarithmic Equations 8.1 Equations Reducible to Quadratic Equations Example 8.6T When x = –3, log x and log (x+1) are undefined, therefore x = –3 is rejected. Solution:

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More About Equations 8 Home Content P. 7 8.2 Problems Leading to Quadratic Equations 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 original problem – not just for the equation found in Step 3.

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More About Equations 8 Home Content P. 8 8.2 Problems Leading to Quadratic Equations Example 8.8T Consider a rectangle with an area of 100 cm 2. 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: 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.

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More About Equations 8 Home Content P. 9 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 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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More About Equations 8 Home Content P. 10 A. Solving Simultaneous Equations by Graphical Method 8.4 Graphical Solutions of Simultaneous Equations 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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More About Equations 8 Home Content P. 11 B. Number of Points of Intersection of a Parabola and a Line 8.4 Graphical Solutions of Simultaneous Equations To solve a pair of simultaneous equations in which one is linear and the other is quadratic (in the form y = ax 2 + 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; or Case 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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More About Equations 8 Home Content P. 12 8.4 Graphical Solutions of Simultaneous Equations Without the actual drawing of the graphs, the number of points of intersection of 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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More About Equations 8 Home Content P. 13 8.4 Graphical Solutions of Simultaneous Equations Example 8.19T Without solving the simultaneous equations algebraically, find the number of points of intersection of the parabola y = 2x 2 and the straight line y = 3x + 5. Δ> 0 corresponds to the quadratic equation 2x 2 – 3x – 5 having two unequal real roots. Solution: Substituting (2) into (1), There are two points of intersection.

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More About Equations 8 Home Content P. 14 8.5 More about Graphical Methods in Solving Simultaneous Equations When we are given a graph of quadratic function such as y = x 2, we can use it to solve any quadratic equation graphically such as x 2 – x – 2 = 0 by the following procedures: Step 1 :Write the equation as x 2 = x + 2. Step 2 :Hence, we can write this quadratic equation as two simultaneous equations ( one linear and one quadratic ) in two unknowns x and y, namely y = x 2 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 points of intersection will give the solutions of the quadratic equation x 2 – x – 2 = 0.

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