Here are the detailed step-by-step solutions to the questions.
1. Move the constant term to the right side: $x^2 + 6x = 7$
2. Halve the coefficient of $x$ (which is 6), square it ($3^2 = 9$), and add it to both sides: $x^2 + 6x + 9 = 7 + 9$
3. Factor the left side as a perfect square: $(x+3)^2 = 16$
4. Take the square root of both sides: $x+3 = \pm\sqrt{16} \implies x+3 = \pm4$
5. Solve for $x$: $x = -3 \pm 4$. This gives two solutions: $x = -3 + 4 = 1$ and $x = -3 - 4 = -7$.
Answer: $x = 1$ or $x = -7$
1. Move the constant term: $x^2 - 8x = -10$
2. Halve the coefficient of $x$ (which is -8), square it ($(-4)^2 = 16$), and add it to both sides: $x^2 - 8x + 16 = -10 + 16$
3. Factor the left side: $(x-4)^2 = 6$
4. Take the square root of both sides: $x-4 = \pm\sqrt{6}$
5. Solve for $x$: $x = 4 \pm\sqrt{6}$.
Answer: $x = 4 + \sqrt{6}$ or $x = 4 - \sqrt{6}$
1. Move the constant term: $x^2 + 5x = 3$
2. Halve the coefficient of $x$ (which is 5), square it ($(5/2)^2 = 25/4$), and add it to both sides: $x^2 + 5x + \frac{25}{4} = 3 + \frac{25}{4}$
3. Factor the left side: $(x + \frac{5}{2})^2 = \frac{12}{4} + \frac{25}{4} = \frac{37}{4}$
4. Take the square root of both sides: $x + \frac{5}{2} = \pm\sqrt{\frac{37}{4}} = \pm\frac{\sqrt{37}}{2}$
5. Solve for $x$: $x = -\frac{5}{2} \pm \frac{\sqrt{37}}{2}$.
Answer: $x = \frac{-5 \pm \sqrt{37}}{2}$
1. The left side is already a perfect square trinomial.
2. Factor the left side: $(x-6)^2 = 0$
3. Take the square root of both sides: $x-6 = 0$
4. Solve for $x$: $x=6$.
Answer: $x = 6$ (a repeated root)
1. The coefficient of $x^2$ is not 1. Divide the entire equation by 2: $x^2 + 6x - 2 = 0$
2. Move the constant term: $x^2 + 6x = 2$
3. Halve the coefficient of $x$ (which is 6), square it ($3^2 = 9$), and add it to both sides: $x^2 + 6x + 9 = 2 + 9$
4. Factor the left side: $(x+3)^2 = 11$
5. Take the square root of both sides: $x+3 = \pm\sqrt{11}$
6. Solve for $x$: $x = -3 \pm \sqrt{11}$.
Answer: $x = -3 + \sqrt{11}$ or $x = -3 - \sqrt{11}$
1. Divide the entire equation by 3: $x^2 - 2x - \frac{5}{3} = 0$
2. Move the constant term: $x^2 - 2x = \frac{5}{3}$
3. Halve the coefficient of $x$ (which is -2), square it ($(-1)^2 = 1$), and add it to both sides: $x^2 - 2x + 1 = \frac{5}{3} + 1$
4. Factor the left side: $(x-1)^2 = \frac{5}{3} + \frac{3}{3} = \frac{8}{3}$
5. Take the square root of both sides: $x-1 = \pm\sqrt{\frac{8}{3}} = \pm\frac{\sqrt{8}}{\sqrt{3}} = \pm\frac{2\sqrt{2}}{\sqrt{3}}$. Rationalise the denominator: $\pm\frac{2\sqrt{6}}{3}$
6. Solve for $x$: $x = 1 \pm \frac{2\sqrt{6}}{3}$.
Answer: $x = 1 + \frac{2\sqrt{6}}{3}$ or $x = 1 - \frac{2\sqrt{6}}{3}$
1. Move the constant term: $x^2 - 3x = 1$
2. Halve the coefficient of $x$ (which is -3), square it ($(-3/2)^2 = 9/4$), and add it to both sides: $x^2 - 3x + \frac{9}{4} = 1 + \frac{9}{4}$
3. Factor the left side: $(x - \frac{3}{2})^2 = \frac{4}{4} + \frac{9}{4} = \frac{13}{4}$
4. Take the square root of both sides: $x - \frac{3}{2} = \pm\sqrt{\frac{13}{4}} = \pm\frac{\sqrt{13}}{2}$
5. Solve for $x$: $x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$.
Answer: $x = \frac{3 \pm \sqrt{13}}{2}$
1. Divide the entire equation by 5: $x^2 + 2x + \frac{1}{5} = 0$
2. Move the constant term: $x^2 + 2x = -\frac{1}{5}$
3. Halve the coefficient of $x$ (which is 2), square it ($1^2=1$), and add it to both sides: $x^2 + 2x + 1 = -\frac{1}{5} + 1$
4. Factor the left side: $(x+1)^2 = \frac{4}{5}$
5. Take the square root of both sides: $x+1 = \pm\sqrt{\frac{4}{5}} = \pm\frac{2}{\sqrt{5}}$. Rationalise the denominator: $\pm\frac{2\sqrt{5}}{5}$
6. Solve for $x$: $x = -1 \pm \frac{2\sqrt{5}}{5}$.
Answer: $x = -1 + \frac{2\sqrt{5}}{5}$ or $x = -1 - \frac{2\sqrt{5}}{5}$
1. Move the constant term: $x^2 + 2x = -5$
2. Halve the coefficient of $x$ (which is 2), square it ($1^2=1$), and add it to both sides: $x^2 + 2x + 1 = -5 + 1$
3. Factor the left side: $(x+1)^2 = -4$
4. The square of a real number cannot be negative. Therefore, there are no real solutions.
Answer: No real solutions.
1. Divide the entire equation by 2: $x^2 - \frac{5}{2}x - 2 = 0$
2. Move the constant term: $x^2 - \frac{5}{2}x = 2$
3. Halve the coefficient of $x$ (which is -5/2), giving -5/4. Square it ($(-5/4)^2 = 25/16$), and add it to both sides: $x^2 - \frac{5}{2}x + \frac{25}{16} = 2 + \frac{25}{16}$
4. Factor the left side: $(x - \frac{5}{4})^2 = \frac{32}{16} + \frac{25}{16} = \frac{57}{16}$
5. Take the square root of both sides: $x - \frac{5}{4} = \pm\sqrt{\frac{57}{16}} = \pm\frac{\sqrt{57}}{4}$
6. Solve for $x$: $x = \frac{5}{4} \pm \frac{\sqrt{57}}{4}$.
Answer: $x = \frac{5 \pm \sqrt{57}}{4}$