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