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WHAT IS A “SOLUTION”? Sect P.5 solving Equations.

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Presentation on theme: "WHAT IS A “SOLUTION”? Sect P.5 solving Equations."— Presentation transcript:

1 WHAT IS A “SOLUTION”? Sect P.5 solving Equations

2 Solution…. Every value of the “variable” that makes the equation true… Is x=3 a solution to x 2 + 5x – 3 = 21? Is (4,7) a solution to 5x – 3y = 1?

3 Quadratic Equations A quadratic equation is an equation of degree 2 … The “degree” is the highest exponent of the variable x 2 + 4x – 7 = 0 is a quadratic equation. 3x – 5 = 8 is not.

4 Completing the Square Isolate the variables on one side Add “some” quantity to each side to make the left side a perfect square Example: if on the left side you have x 2 + 4x then you should add 4 So that the left side becomes 4

5 What do you add? StartAddResult X 2 + 12x36X 2 + 12x + 36 = (x + 6) 2 X 2 – 6x9X 2 – 6x + 9 = (x – 3) 2 X 2 + x¼X 2 + x + ¼ = (x + ½) 2 Can you guess the rule? 5 StartAddResult X + mx(m/2) 2 (x + mx + (m/2) 2 = (x + m/2) 2

6 Try this one… Solve x 2 +4x -21 = 0 6

7 Quadratic Formula Standard Form of the Quadratic Equation ax 2 + bx + c = 0 Solutions are: There are “real” solutions if b 2 -4ac is positive … That is, 7

8 Solving by Quadratic Formula 1. Check to see if b 2 – 4ac > 0… if not, there are no real solutions 2. Substitute values of a, b, c into the formula and simplify Solve 2x 2 – 3x – 1 = 0 8

9 Quadratic Equations Factoring  Look for Common Factors first  Put the equation in form ax 2 + bx +c = 0  Factor the left side of the equation into the product of two binomials  Set each factor equal to 0 … the zero-product rule  Example Solve x 2 = 4x 9

10 Factoring x 2 - x – 6 = 0 2x 3 + 2x 2 – 12x = 0 10

11 Square Root Method Take the square of both sides.. Remember, there will be a positive and a negative solution Also remember Solve x 2 = 5 Solve (x – 2) 2 = 16 11

12 Solving by Calculator Solve for y= Enter into y= screen on calculator Graph Method 1 … use “trace” to find where the point where y=0 Method 2… use “table” to find where y = 0 Method 3 … put left side in as y1 and right side as y2  Use “intersect” to find the intersection point… the x value of the intersection point is the solution.


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