Question

3x^2-15=0, x^2+7x=0, 12x^2-5x-2=0

Answer 1

$3 {x}^{2} - 15 = 0 \\ {x}^{2} - 5 = 0 \\ x = + - \sqrt{5}$

${x}^{2} + 7x = 0 \\ x(x + 7) = 0 \\ x = 0 \\ x = - 7$

$12 {x}^{2} - 5x - 2 = 0$
D = 25+96=121
$x = \frac{5 + 11}{2 \times 12 } = \frac{2}{3} \\ x = \frac{5 - 11}{2 \times 12} = - \frac{1}{4}$

Answer 2

To solve each of these quadratic equations, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each equation step by step:

1. 3x^2 - 15 = 0

First, we'll bring all terms to one side of the equation:

3x^2 - 15 = 0 3x^2 = 15 x^2 = 15 / 3 x^2 = 5

Now, we'll take the square root of both sides:

x = ±√5

So the solutions are x = √5 and x = -√5.

2. x^2 + 7x = 0

Again, bring all terms to one side:

x^2 + 7x = 0

Factor out x:

x(x + 7) = 0

Now, apply the zero product property (if the product of two factors is zero, then at least one of the factors must be zero):

x = 0 or x + 7 = 0

Solving for x:

x = 0 or x = -7

So the solutions are x = 0 and x = -7.

3. 12x^2 - 5x - 2 = 0

We'll use the quadratic formula for this equation:

a = 12, b = -5, and c = -2

x = (-(-5) ± √((-5)^2 - 4 * 12 * (-2))) / (2 * 12) x = (5 ± √(25 + 96)) / 24 x = (5 ± √121) / 24 x = (5 ± 11) / 24

Now, we have two possibilities:

1. x = (5 + 11) / 24 x = 16 / 24 x = 2/3

2. x = (5 - 11) / 24 x = -6 / 24 x = -1/4

So the solutions are x = 2/3 and x = -1/4.

Please note that these solutions are approximate decimals; if you prefer exact values, you can leave them as square roots or fractions.