Math worksheet with 10 quadratic equations to solve using the sum of perfect squares method.

It looks like you've uploaded an image containing a set of quadratic equations in the form $(x + a)^2 = b$. These are perfect square equations, and we can solve each one by taking the square root of both sides.

Let’s go through each equation step-by-step and solve for $x$.

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Step-by-step Solutions:

We use the rule:
> If $(x + a)^2 = b$, then $x + a = \pm\sqrt{b}$, so $x = -a \pm \sqrt{b}$

But only real solutions exist if $b \geq 0$. If $b < 0$, there are no real solutions.

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#### 1. $(x+2)^2 = 36$
$$
x + 2 = \pm \sqrt{36} = \pm 6 \\
x = -2 \pm 6 \\
x = -2 + 6 = 4 \quad \text{or} \quad x = -2 - 6 = -8
$$
✔ Solutions: $x = 4, -8$

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#### 2. $(x-1)^2 = 16$
$$
x - 1 = \pm \sqrt{16} = \pm 4 \\
x = 1 \pm 4 \\
x = 5 \quad \text{or} \quad x = -3
$$
✔ Solutions: $x = 5, -3$

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#### 3. $(x+1)^2 = 36$
$$
x + 1 = \pm \sqrt{36} = \pm 6 \\
x = -1 \pm 6 \\
x = 5 \quad \text{or} \quad x = -7
$$
✔ Solutions: $x = 5, -7$

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#### 4. $(x+2)^2 = 64$
$$
x + 2 = \pm \sqrt{64} = \pm 8 \\
x = -2 \pm 8 \\
x = 6 \quad \text{or} \quad x = -10
$$
✔ Solutions: $x = 6, -10$

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#### 5. $(x-2)^2 = 16$
$$
x - 2 = \pm \sqrt{16} = \pm 4 \\
x = 2 \pm 4 \\
x = 6 \quad \text{or} \quad x = -2
$$
✔ Solutions: $x = 6, -2$

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#### 6. $(x-1)^2 = 4$
$$
x - 1 = \pm \sqrt{4} = \pm 2 \\
x = 1 \pm 2 \\
x = 3 \quad \text{or} \quad x = -1
$$
✔ Solutions: $x = 3, -1$

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#### 7. $(x+4)^2 = 1$
$$
x + 4 = \pm \sqrt{1} = \pm 1 \\
x = -4 \pm 1 \\
x = -3 \quad \text{or} \quad x = -5
$$
✔ Solutions: $x = -3, -5$

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#### 8. $(x+4)^2 = 4$
$$
x + 4 = \pm \sqrt{4} = \pm 2 \\
x = -4 \pm 2 \\
x = -2 \quad \text{or} \quad x = -6
$$
✔ Solutions: $x = -2, -6$

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#### 9. $(x-1)^2 = -2$
$$
\text{Square root of negative number → No real solution}
$$
✘ No real solutions (only complex: $x = 1 \pm i\sqrt{2}$)

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#### 10. $(x+2)^2 = 16$
$$
x + 2 = \pm \sqrt{16} = \pm 4 \\
x = -2 \pm 4 \\
x = 2 \quad \text{or} \quad x = -6
$$
✔ Solutions: $x = 2, -6$

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#### 11. $(x+1)^2 = 9$
$$
x + 1 = \pm \sqrt{9} = \pm 3 \\
x = -1 \pm 3 \\
x = 2 \quad \text{or} \quad x = -4
$$
✔ Solutions: $x = 2, -4$

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#### 12. $(x+3)^2 = 4$
$$
x + 3 = \pm \sqrt{4} = \pm 2 \\
x = -3 \pm 2 \\
x = -1 \quad \text{or} \quad x = -5
$$
✔ Solutions: $x = -1, -5$

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#### 13. $(x-2)^2 = 1$
$$
x - 2 = \pm \sqrt{1} = \pm 1 \\
x = 2 \pm 1 \\
x = 3 \quad \text{or} \quad x = 1
$$
✔ Solutions: $x = 3, 1$

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#### 14. $(x-3)^2 = 16$
$$
x - 3 = \pm \sqrt{16} = \pm 4 \\
x = 3 \pm 4 \\
x = 7 \quad \text{or} \quad x = -1
$$
✔ Solutions: $x = 7, -1$

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✔ Final Summary of All Solutions:

| Equation | Solution(s) |
|--------|-------------|
| $(x+2)^2 = 36$ | $x = 4, -8$ |
| $(x-1)^2 = 16$ | $x = 5, -3$ |
| $(x+1)^2 = 36$ | $x = 5, -7$ |
| $(x+2)^2 = 64$ | $x = 6, -10$ |
| $(x-2)^2 = 16$ | $x = 6, -2$ |
| $(x-1)^2 = 4$ | $x = 3, -1$ |
| $(x+4)^2 = 1$ | $x = -3, -5$ |
| $(x+4)^2 = 4$ | $x = -2, -6$ |
| $(x-1)^2 = -2$ | No real solution |
| $(x+2)^2 = 16$ | $x = 2, -6$ |
| $(x+1)^2 = 9$ | $x = 2, -4$ |
| $(x+3)^2 = 4$ | $x = -1, -5$ |
| $(x-2)^2 = 1$ | $x = 3, 1$ |
| $(x-3)^2 = 16$ | $x = 7, -1$ |

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Let me know if you'd like these organized in a table or need help graphing or checking any of them!