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4 years ago

Solve the Quadratics:

Mathematics

1 answer:

Inessa [10]4 years ago

4 0

Answer: 1. {-2, -3} 2. {-5, 5} 3. {-3, 5}

<u>Step-by-step explanation:</u>

1) First, factor the equation by finding two numbers whose product is 6 and sum is 5. Then apply the Zero Product Property by setting each product equal to zero and solving for m.

m² + 5m + 6 = 0

∧

1 + 6 = 7

2 + 3 = 5 This works!

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

x + 2 = 0 x + 3 = 0

x = -2 x = -3

2) Factor out the GCF of 5. Notice the remaining factor is the difference of squares (because the middle term is missing and the first and last terms are perfect squares. Then apply the Zero Product Property by setting each product equal to zero and solving for p.

5p² - 125 = 0

5(p² - 25) = 0

5(p +5)(p - 5) = 0

5 ≠ 0 p+ 5 = 0 p - 5 = 0

p = -5 p = 5

3) Factor out the GCF of 2. Factor the equation by finding two numbers whose product is -15 and sum is -2. Then apply the Zero Product Property by setting each product equal to zero and solving for x.

2x² - 4x - 30 = 0

2(x² - 2 - 15) = 0

∧

1 - 15 = -14

3 - 5 = -2 This works!

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

x + 3 = 0 x - 5 = 0

x = -3 x = 5

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A quadratic equation is modeled by:

$y = ax^2 + bx + c$

The discriminant is:

$\Delta = b^2 - 4ac$

The solutions are as follows:

If $\mathbf{\Delta > 0}$ , it has 2 real solutions.

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If $\mathbf{\Delta < 0}$ , it has 0 real solutions.

Hence, for the given values of the discriminant, we have that:

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discriminant = 3: 2 Real number solutions.

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More can be learned about quadratic functions at brainly.com/question/24737967

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