2a

2a We get, x =

2(1) x =

2 x =

2 x =

2 Now, we can say, x =

2 x =

2 x = -2

2 x =

2 x = -5 So, your answer is x = -2 and x = -5.

2a We get, x =

2(5) x =

10 x =

10 x =

10 Now, we can say, x =

10 x =

10 x = -0.4

10 x =

10 x = -2 So, your answer is x = -0.4 and x = -2.

**Quadratic Equations****Before doing this lesson, you should first familarize yourself with the lesson on quadratic expressions.**A quadratic equation is written in the form a^{2}+ bx + c = 0. You can solve a quadratic equation by using various methods. For now, we will be looking at two different methods: 1. Factorization 2. The Quadratic Formula**Method 1: Using the method of factorization**Example 1: Factorize a^{2}– 49 = 0.**Solution:**a^{2}– 49 = 0 can be written as (a - 7)(a + 7) = 0 Either (a - 7 = 0 or a + 7 = 0 By solving these two equations, we get a -7 = 0 a = 7 and a + 7 = 0 so, a = -7. Therefore, our answer is a = 7 and a = -7. You could also write your answer as a = ±7, which is read as "a is equal to plus or minus seven".__Example 2:__Solve x^{2}+ 7x + 12 = 0**Solution:**x^{2}+ 7x + 12 = 0 x^{2}+ 3x + 4x + 12 = 0 x(x + 3) + 4(x + 3) = 0 (x + 3)(x + 4) = 0 Either x + 3 = 0 or x + 4 = 0 By solving these two equations, we get x + 3 = 0 x = -3 and x + 4 = 0 so, x = -4. Therefore, our answer is a = -3 and a = -4.**Method 2: The Quadratic Formula**You must keep in mind that not all quadratic equations can be solved by factorization. The formula is a method that can be used to solved all quadratic equations. The quadratic formula is derived from the general quadratic equation a^{2}+ bx + c = 0. The formula is given as x =__-b ±__√__b__^{2}__- 4ac__2a

__Example 1:__Solve x^{2}+ 7x + 10 = 0. There is no number written in front of the x^{2}term, but in that case it is helpful to think of the x^{2}term as 1x^{2}, so that we have: a = 1, b = 7 and c = 10 By substituting these values into the formula: x =__-b ±__√__b__^{2}__- 4ac__2a We get, x =

__-7 ±__√__7__^{2}__- 4(1)(10)__2(1) x =

__-7 ±__√__49 - 40__2 x =

__-7 ±__√__9__2 x =

__-7 ± 3__2 Now, we can say, x =

__-7 + 3__2 x =

__-4__2 x = -2

**OR**x =__-7 - 3__2 x =

__-10__2 x = -5 So, your answer is x = -2 and x = -5.

__Example 2:__Solve 5x^{2}+ 12x + 4 = 0. In this example, a = 5, b = 12 and c = 4 By substituting these values into the formula: x =__-b ±__√__b__^{2}__- 4ac__2a We get, x =

__-12 ±__√__12__^{2}__- 4(5)(4)__2(5) x =

__-12 ±__√__144 - 80__10 x =

__-12 ±__√__64__10 x =

__-7 ± 8__10 Now, we can say, x =

__-12 + 8__10 x =

__-4__10 x = -0.4

**OR**x =__-12 - 8__10 x =

__-20__10 x = -2 So, your answer is x = -0.4 and x = -2.

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