Certainly! Let us solve the given quadratic equations step-by-step:
### a) [tex]\(3 x^2 + 6 x = 0\)[/tex]
1. Factor out the common term:
[tex]\[
3x(x + 2) = 0
\][/tex]
2. Set each factor to zero:
[tex]\[
3x = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 0 \quad \text{or} \quad x = -2
\][/tex]
So, the solutions are [tex]\(x = -2\)[/tex] and [tex]\(x = 0\)[/tex].
### b) [tex]\(3 x^2 - 27 = 0\)[/tex]
1. Add 27 to both sides:
[tex]\[
3x^2 = 27
\][/tex]
2. Divide by 3:
[tex]\[
x^2 = 9
\][/tex]
3. Take the square root of both sides:
[tex]\[
x = \pm 3
\][/tex]
So, the solutions are [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex].
### c) [tex]\(x^2 - 25 = 0\)[/tex]
1. Add 25 to both sides:
[tex]\[
x^2 = 25
\][/tex]
2. Take the square root of both sides:
[tex]\[
x = \pm 25
\][/tex]
So, the solutions are [tex]\(x = 5\)[/tex] and [tex]\(x = -5\)[/tex].
### d) [tex]\(2 x^2 + x = 0\)[/tex]
1. Factor out the common term:
[tex]\[
x(2x + 1) = 0
\][/tex]
2. Set each factor to zero:
[tex]\[
x = 0 \quad \text{or} \quad 2x + 1 = 0
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 0 \quad \text{or} \quad 2x = -1 \quad \text{so} \quad x = -\frac{1}{2}
\][/tex]
So, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{1}{2}\)[/tex].
### e) [tex]\(4 x^2 - 9 = 0\)[/tex]
1. Add 9 to both sides:
[tex]\[
4x^2 = 9
\][/tex]
2. Divide by 4:
[tex]\[
x^2 = \frac{9}{4}
\][/tex]
3. Take the square root of both sides:
[tex]\[
x = \pm \frac{3}{2}
\][/tex]
So, the solutions are [tex]\(x = \frac{3}{2}\)[/tex] and [tex]\(x = -\frac{3}{2}\)[/tex].
### f) [tex]\(5 x^2 - 10 x = 0\)[/tex]
1. Factor out the common term:
[tex]\[
5x(x - 2) = 0
\][/tex]
2. Set each factor to zero:
[tex]\[
5x = 0 \quad \text{or} \quad x - 2 = 0
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 0 \quad \text{or} \quad x = 2
\][/tex]
So, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = 2\)[/tex].
