Question

SHOW ALL THE STEPS


A. Solve by using graphing
⬇️
1. 3x²=6x-3


2. 4x²-12x+9=0


B. Solve by using square root
⬇️
3. 2(x+2)²- 5=9


4. -2x²+=-6

C. Solve by using factoring
⬇️
5. x²-11x= -30


6. f(x)=x²+2x-8

Answer 1

Let's solve each problem step by step.

### A. Solve by using graphing

**1. \(3x²=6x-3\)**

First, rearrange the equation to set it to 0:
\[
3x^2 - 6x + 3 = 0
\]
Now, we can graph the function \(f(x) = 3x^2 - 6x + 3\) to find the roots:

Identify the vertex and x-intercepts:
- Vertex: Use the formula \(x = -\frac{b}{2a} = -\frac{-6}{2(3)} = 1\).
- Substitute \(x = 1\) back into the function to find the y-coordinate of the vertex:
\[
f(1) = 3(1)^2 - 6(1) + 3 = 3 - 6 + 3 = 0
\]
The vertex is (1,0). Since the parabola opens upward (a > 0), the graph crosses the x-axis at x = 1.

By plotting the graph, you can approximate where it intersects the x-axis. Since we calculated one root directly, we find:
\[
x = 1 \quad (\text{double root})
\]

**2. \(4x² - 12x + 9 = 0\)**

Next, we graph the function \(g(x) = 4x^2 - 12x + 9\):

This can also be factored:
\[
(2x - 3)^2 = 0
\]
So the roots are:
\[
2x - 3 = 0 \Rightarrow x = \frac{3}{2}
\]

Both methods lead us to the same result. Plot the quadratic, which will touch the x-axis at \(x = \frac{3}{2}\).

### B. Solve by using square root

**3. \(2(x + 2)^2 - 5 = 9\)**

First, isolate the square term:
\[
2(x + 2)^2 = 14 \\
(x + 2)^2 = 7
\]
Now, take the square root:
\[
x + 2 = \pm \sqrt{7}
\]
Solving for \(x\):
\[
x = -2 \pm \sqrt{7}
\]
Thus, the solutions are:
\[
x = -2 + \sqrt{7} \quad \text{and} \quad x = -2 - \sqrt{7}
\]

**4. \(-2x² = -6\)**

Rearranging gives:
\[
2x^2 = 6 \\
x^2 = 3
\]
Taking the square root:
\[
x = \pm \sqrt{3}
\]
The solutions are:
\[
x = \sqrt{3} \quad \text{and} \quad x = -\sqrt{3}
\]

### C. Solve by using factoring

**5. \(x² - 11x = -30\)**

Rearrange the equation:
\[
x^2 - 11x + 30 = 0
\]
Factoring gives:
\[
(x - 5)(x - 6) = 0
\]
Thus, setting each factor to zero:
\[
x - 5 = 0 \quad \Rightarrow \quad x = 5 \\
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]
So the solutions are:
\[
x = 5 \quad \text{and} \quad x = 6
\]

**6. \(f(x) = x² + 2x - 8\)**

Set the function equal to zero to solve for the roots:
\[
x^2 + 2x - 8 = 0
\]
Factoring:
\[
(x + 4)(x - 2) = 0
\]
Setting the factors equal to zero:
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4 \\
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
The solutions are:
\[
x = -4 \quad \text{and} \quad x = 2
\]

### Summary of Solutions
1. \(x = 1\) (double root)
2. \(x = \frac{3}{2}\)
3. \(x = -2 + \sqrt{7}, -2 - \sqrt{7}\)
4. \(x = \sqrt{3}, -\sqrt{3}\)
5. \(x = 5, 6\)
6. \(x = -4, 2\)