https://file.aiquickdraw.com/mathbot/file/1729167455206anmaq41c.jpg

моля да решиш 2 и 9 задача

Let's solve tasks 2 and 9 from the image.

### Task 2:
\[
\sqrt{x^2 - 12x + 36} = 4
\]

1. First, eliminate the square root by squaring both sides:
   \[
   x^2 - 12x + 36 = 16
   \]

2. Simplify the equation:
   \[
   x^2 - 12x + 36 - 16 = 0
   \]
   \[
   x^2 - 12x + 20 = 0
   \]

3. Solve the quadratic equation using the discriminant:
   \[
   D = (-12)^2 - 4 \cdot 1 \cdot 20 = 144 - 80 = 64
   \]

4. Now, apply the quadratic formula:
   \[
   x = \frac{-(-12) \pm \sqrt{64}}{2 \cdot 1} = \frac{12 \pm 8}{2}
   \]

   So, the solutions are:
   \[
   x_1 = \frac{12 + 8}{2} = 10, \quad x_2 = \frac{12 - 8}{2} = 2
   \]

5. Check for extraneous solutions by substituting back into the original equation:
   - For \(x = 10\):
     \[
     \sqrt{10^2 - 12 \cdot 10 + 36} = \sqrt{100 - 120 + 36} = \sqrt{16} = 4
     \]
     This is correct.

   - For \(x = 2\):
     \[
     \sqrt{2^2 - 12 \cdot 2 + 36} = \sqrt{4 - 24 + 36} = \sqrt{16} = 4
     \]
     This is also correct.

Thus, the solutions are:
\[
x = 10 \quad \text{or} \quad x = 2
\]

---

### Task 9:
\[
\sqrt{3x + 13} - \sqrt{2x + 7} = 1
\]

1. Isolate one square root, for example:
   \[
   \sqrt{3x + 13} = \sqrt{2x + 7} + 1
   \]

2. Square both sides:
   \[
   3x + 13 = (\sqrt{2x + 7} + 1)^2
   \]
   Expanding the right-hand side:
   \[
   3x + 13 = (2x + 7) + 2\sqrt{2x + 7} + 1
   \]
   \[
   3x + 13 = 2x + 8 + 2\sqrt{2x + 7}
   \]

3. Simplify:
   \[
   3x + 13 - 2x - 8 = 2\sqrt{2x + 7}
   \]
   \[
   x + 5 = 2\sqrt{2x + 7}
   \]

4. Isolate the square root again:
   \[
   \frac{x + 5}{2} = \sqrt{2x + 7}
   \]

5. Square both sides:
   \[
   \left(\frac{x + 5}{2}\right)^2 = 2x + 7
   \]
   Expanding:
   \[
   \frac{(x + 5)^2}{4} = 2x + 7
   \]
   Multiply both sides by 4:
   \[
   (x + 5)^2 = 8x + 28
   \]
   Expanding:
   \[
   x^2 + 10x + 25 = 8x + 28
   \]

6. Simplify and form a quadratic equation:
   \[
   x^2 + 10x + 25 - 8x - 28 = 0
   \]
   \[
   x^2 + 2x - 3 = 0
   \]

7. Solve the quadratic equation using the discriminant:
   \[
   D = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16
   \]
   \[
   x = \frac{-2 \pm \sqrt{16}}{2 \cdot 1} = \frac{-2 \pm 4}{2}
   \]
   The solutions are:
   \[
   x_1 = \frac{-2 + 4}{2} = 1, \quad x_2 = \frac{-2 - 4}{2} = -3
   \]

8. Check for extraneous solutions:
   - For \(x = 1\):
     \[
     \sqrt{3 \cdot 1 + 13} - \sqrt{2 \cdot 1 + 7} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1
     \]
     This is correct.

   - For \(x = -3\):
     \[
     \sqrt{3 \cdot (-3) + 13} - \sqrt{2 \cdot (-3) + 7} = \sqrt{-9 + 13} - \sqrt{-6 + 7} = \sqrt{4} - \sqrt{1} = 2 - 1 = 1
     \]
     This is also correct.

Thus, the solutions are:
\[
x = 1 \quad \text{or} \quad x = -3
\]

---

Let me know if you'd like more details or have questions! Here are 5 related questions for further practice:

1. How do you solve quadratic equations using the quadratic formula?
2. What are the steps to isolate a square root in an equation?
3. Why do we need to check for extraneous solutions in radical equations?
4. What are other methods for solving quadratic equations besides the quadratic formula?
5. How can you simplify radical expressions involving variables?

**Tip:** Always check for extraneous solutions when dealing with equations involving square roots, as squaring both sides can introduce non-valid solutions.

Learn how to solve radical equations from tasks 2 and 9, involving quadratic equations and radical simplifications. This step-by-step guide helps students grasp the methods to solve complex algebraic equations, suitable for Grades 8-10.

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