Answer:
1. x² + 4x = 0
To solve this equation, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
Where:
- a = 1
- b = 4
- c = 0
Plugging in the values, we get:
x = (-4 ± √(4² - 4(1)(0))) / (2(1))
x = (-4 ± 4) / 2
x = 0 or -2
Therefore, the solutions to the equation x² + 4x = 0 are x = 0 and x = -2.
2. 3x² - 15 = 0
To solve this equation, we can divide both sides by 3 to get:
x² - 5 = 0
Now, we can use the quadratic formula again:
x = (-b ± √(b² - 4ac)) / (2a)
Where:
- a = 1
- b = 0
- c = -5
Plugging in the values, we get:
x = (0 ± √(0² - 4(1)(-5))) / (2(1))
x = ±√20 / 2
x = ±2√5 / 2
x = ±√5
Therefore, the solutions to the equation 3x² - 15 = 0 are x = √5 and x = -√5.
3. (x + 3)(x - 7) = 0
This equation can be solved by finding the values of x that make either (x + 3) or (x - 7) equal to 0.
Setting (x + 3) = 0, we get:
x = -3
Setting (x - 7) = 0, we get:
x = 7
Therefore, the solutions to the equation (x + 3)(x - 7) = 0 are x = -3 and x = 7.
Answer:
1. x=0,-4
2. x=[tex]\sqrt{5}[/tex]
3. x=-3,7
Step-by-step explanation:
1. x2+4x=0
x(x+4)=0
x=0
x+4=0
x= -4
2. 3x2-15=0
3x2=15
x2=15
x= [tex]\sqrt{5}[/tex]
3. x+3=0
x= -3
x-7=0
x= 7
