solve equations

Question

1. 3x-5=0 
2. x²+4=0 
3. x²+2x+1=0 
4. x³+2x²-15x=0 

Answer 1

1) The linear equation 3x - 5 = 0

Add 5 to each side.

3x - 5 + 5 = 0 + 5

3x = 5

Divide each side by 3.

3x/3 = 5/3

Solution x = 5.

Answer 2

2) The equation x² + 4 = 0

Subtacct 4 from each side.

x²  = - 4

Apply square root on each side.

√x²  = √- 4

√x²  = √[(i²)(4)]

x = ± 2i

Solutions are 2i and - 2i.

Answer 3

3) The quadratic equation x² + 2x + 1 = 0

Recall the formula (a + b)² = a² + b² + 2ab

Rewrite the equation (x)² + (2)(x)(1) + (1)² = 0

In this case a = x, b = 1

(x + 1)² = 0

Apply square root on each side.

x + 1 = 0

Solution x = - 1.

Answer 4

4) The quadratic equation x³ + 2x² - 15x = 0

x( x² + 2x - 15) = 0

Apply zero product property.

x = 0 and x² + 2x - 15 = 0

x = 0 is one solution.

Solve the equation x² + 2x - 15 = 0 by factorization.

Multiply first term x² and last term -15 = - 15x²

The correct pair of the terms 5x and -3x multiply to -15x² and add to 2x.

Replace the middle term 2x with 5x - 3x.

 x² + 5x - 3x - 15 = 0

Group the terms into two pairs.

(x² + 5x) + (- 3x - 15) = 0

Factor out x from the first group  and factor out -3 from the second group.

x(x + 5) - 3(x + 5) = 0

Factor out common term x + 5.

 (x + 5)(x - 3) = 0

Apply zero product property.

x + 5 = 0 and x - 3 = 0

x = - 5 and x = 3

Solutions of x³ + 2x² - 15x = 0 are at x = 0, -5, 3.