Question

Solve the equation
1. 4x^2-4x-24=0
2. 3p^2-5p-28=0

Answer 1

To solve the equation 4x^2-4x-24=0, we can start by factoring out a 4:
4(x^2-x-6) = 0

Next, we can factor the quadratic expression in the parentheses:

4(x-3)(x+2) = 0

This gives us two solutions:

x-3 = 0, so x = 3
x+2 = 0, so x = -2

Therefore, the solutions to the equation 4x^2-4x-24=0 are x=3 and x=-2.

To solve the equation 3p^2-5p-28=0, we can use the quadratic formula:
p = (-b ± sqrt(b^2-4ac)) / 2a

where a = 3, b = -5, and c = -28. Substituting these values, we get:

p = (-(-5) ± sqrt((-5)^2-4(3)(-28))) / 2(3)

Simplifying:

p = (5 ± sqrt(289)) / 6

p = (5 ± 17) / 6

Therefore, the solutions to the equation 3p^2-5p-28=0 are p = 6 and p = -4/3.

Answer 2

Answer:

1. x = 3 and x = -2.
2. p = 4/3 and p = -7/3.

Explanation:

4x^2-4x-24=0
First, we need to factor out the greatest common factor, which is 4:

4(x^2 - x - 6) = 0

Next, we can factor the quadratic expression inside the parentheses:

4(x - 3)(x + 2) = 0

Now we can use the zero product property, which states that if the product of two factors is 0, then at least one of the factors must be 0. So we set each factor equal to 0 and solve for x:

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

Therefore, the solutions to the equation 4x^2-4x-24=0 are x = 3 and x = -2.

3p^2-5p-28=0
To solve this quadratic equation, we can use the quadratic formula:

p = (-b ± √(b^2 - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In this case, a = 3, b = -5, and c = -28. So we have:

p = (-(-5) ± √((-5)^2 - 4(3)(-28))) / (2(3))

p = (5 ± √(25 + 336)) / 6

p = (5 ± √361) / 6

p = (5 ± 19) / 6

p = 4/3 or p = -7/3

Therefore, the solutions to the equation 3p^2-5p-28=0 are p = 4/3 and p = -7/3.