Let's solve the equations as per your request.
A. Solve by using graphing
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\(3x^2 = 6x - 3\)
Rearranging the equation gives: \[ 3x^2 - 6x + 3 = 0 \] Dividing through by 3: \[ x^2 - 2x + 1 = 0 \] This can be factored as: \[ (x - 1)^2 = 0 \] Thus, the solution is: \[ x = 1 \]
(Note: The graph of \(y = 3x^2\) and \(y = 6x - 3\) will intersect at the point \( (1, 0) \).)
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\(4x^2 - 12x + 9 = 0\)
This can be factored: \[ (2x - 3)^2 = 0 \] Thus, the solution is: \[ x = \frac{3}{2} \text{ (or } x = 1.5\text{)} \]
B. Solve by using square root
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\(2(x+2)^2 - 5 = 9\)
Rearranging gives: \[ 2(x+2)^2 = 14 \implies (x+2)^2 = 7 \] Taking the square root: \[ x + 2 = ±\sqrt{7} \implies x = -2 ± \sqrt{7} \] Thus, the solutions are: \[ x = -2 + \sqrt{7} \quad \text{and} \quad x = -2 - \sqrt{7} \]
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\(-2x^2 = -6\)
Dividing through by -2 gives: \[ x^2 = 3 \] Taking the square root: \[ x = ±\sqrt{3} \] Thus, the solutions are: \[ x = \sqrt{3} \quad \text{and} \quad x = -\sqrt{3} \]
C. Solve by using factoring
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\(x^2 - 11x = -30\)
Rearranging gives: \[ x^2 - 11x + 30 = 0 \] This can be factored as: \[ (x - 5)(x - 6) = 0 \] Thus, the solutions are: \[ x = 5 \quad \text{and} \quad x = 6 \]
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\(f(x) = x^2 + 2x - 8\)
We need to find the zeros (or roots) of the function: \[ x^2 + 2x - 8 = 0 \] This can be factored as: \[ (x + 4)(x - 2) = 0 \] Thus, the solutions are: \[ x = -4 \quad \text{and} \quad x = 2 \]
Summary of Solutions:
- \(x = 1\)
- \(x = \frac{3}{2}\)
- \(x = -2 + \sqrt{7}, , -2 - \sqrt{7}\)
- \(x = \sqrt{3}, , -\sqrt{3}\)
- \(x = 5, , 6\)
- \(x = -4, , 2\)