2. Solving Equations Solve for a certain variable, which is normally x. Involves re-arranging the equation to get x by itself. E.g. 2x + 6 = 4, 2x = -2 x = -1
3. Solving using the null factor law If the product of any two numbers is zero, then one or both of the numbers is zero.That is, if ab = 0, then a = 0 or b = 0 (which includes the possibility that a = b = 0). E.g if (x-6)(x-3) = 0 then x is said to be 6 or 3. Helps us to find the x intercepts and y intercepts of a graph as the graph will intersect the axis when x or y equal zero.
4. Solving Exponential and Logarithmic Equations a^x = a^y, implies x=y E.g. 3^(x-1) = 256 3^(x-1) = 3^4 x – 1 = 4, x = 5 Sometimes you need to re-arrange the equation back into its logarithmic form, using the rule logax = y if ay = x E.g. Solve the equation 0=-5e(x/2) for x: (Do Question Together as a Class) (Solution Page 157 Notes Book)
6. Solving Logarithmic Functions Cont. E.g. Log2x = 5 x = 25 , x = 32 Logx27 = 3/2 x3/2 = 27 x = 272/3 , x = 9 Loge(x-1) + loge(x+2) = loge(6x-8) loge(x-1)(x+2) = loge(6x-8) x2 + x – 2 = 6x – 8 x2 – 5x + 6 = 0 (x-3)(x-2) = 0, x = 3 or x = 2
7. Book Questions Do Question 4.e, page 175 Complete; Exercise 5.D; Q3.a, c, e, i. Q4. a , b, c,
8. Solving using the Solve( command Enter solve( by pressing F2 and then 1. Next type your equation in, and then put a comma and type in x (or whatever variable your solving for). Then close the brackets. E.g. for Loge(x-1) + loge(x+2) = loge(6x-8) you would type solve(ln(x-1) + ln(x+2) = ln(6x-8),x)
9. Solving Circular Functions Involves re-arranging to get x by itself. Do example 18b, page 214 on the board. Solve this equation individually; solve √2sinx + 1 = 0 for x.
10. Book Questions Complete; Exercise 6.H; Q.1. a, c, e – only finding x interecepts.Q2. a, c, e – only finding x intercepts for the interval [0,2π]
