Solve for \( x \) : (Where applicable, leave your answers as indicated in brackets.) \( \begin{array}{ll}\text { (1) } x^{2}-2 x=3 & \text { (2) } 3 x^{2}-27=0 \\ \text { (3) } x^{2}-3 x-2=0 \text { (surd form) } & \text { (4) }-x^{2}-x=-5 \text { (surd form) } \\ \text { (5) }-x^{2}-5 x=2 \text { (surd form) } & \text { (6) } x^{2}-11 x=-30 \\ \text { (7) } x^{2}-12=0 \text { (surd form) } & \text { (8) }-2 x^{2}+6 x=0\end{array} \)
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Let's solve each equation one by one: **(1)** \( x^{2}-2x=3 \) Rearranging gives us \( x^{2}-2x-3=0 \). Factoring, we have \( (x-3)(x+1)=0 \). Thus, \( x=3 \) or \( x=-1 \). --- **(2)** \( 3x^{2}-27=0 \) Dividing by 3, we get \( x^{2}-9=0 \). Factoring gives \( (x-3)(x+3)=0 \), so \( x=3 \) or \( x=-3 \). --- **(3)** \( x^{2}-3x-2=0 \) Using the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), we get \( x=\frac{3\pm\sqrt{(3)^{2}-4(1)(-2)}}{2(1)} = \frac{3\pm\sqrt{17}}{2} \). --- **(4)** \( -x^{2}-x=-5 \) Rearranging gives \( x^{2}+x-5=0 \). Using the quadratic formula: \( x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}}{2(1)}= \frac{-1\pm\sqrt{21}}{2} \). --- **(5)** \( -x^{2}-5x=2 \) Rearranging gives \( x^{2}+5x+2=0 \). Using the quadratic formula: \( x=\frac{-5\pm\sqrt{5^{2}-4(1)(2)}}{2(1)}= \frac{-5\pm\sqrt{17}}{2} \). --- **(6)** \( x^{2}-11x=-30 \) Rearranging gives \( x^{2}-11x+30=0 \). Factoring gives \( (x-6)(x-5)=0 \), so \( x=6 \) or \( x=5 \). --- **(7)** \( x^{2}-12=0 \) This can be rearranged to \( x^{2}=12 \) leading to \( x=\pm 2\sqrt{3} \). --- **(8)** \( -2x^{2}+6x=0 \) Factoring out \(-2x\) gives \( -2x(x-3)=0 \), hence \( x=0 \) or \( x=3 \). --- In summary, the answers for the equations are: 1. \( x = 3, -1 \) 2. \( x = 3, -3 \) 3. \( x = \frac{3 \pm \sqrt{17}}{2} \) 4. \( x = \frac{-1 \pm \sqrt{21}}{2} \) 5. \( x = \frac{-5 \pm \sqrt{17}}{2} \) 6. \( x = 5, 6 \) 7. \( x = \pm 2\sqrt{3} \) 8. \( x = 0, 3 \)
