solve the following equations.
17. ( x^2 - 9 = 0 )
20. ( (x + 4)^2 = 25 )
23. ( (x - 7)^2 = -8 )
26. ( x^2 - 6x + 13 = 0 )
29. ( 2x^2 - x + 30 = 0 )

Question

Accepted Answer

### Problem 17: $ x^2 - 9 = 0 $
# Explanation:
## Step 1: Rewrite as difference of squares
We can rewrite the equation $ x^2 - 9 = 0 $ as a difference of squares: $ x^2 - 3^2 = 0 $. The formula for the difference of squares is $ a^2 - b^2=(a - b)(a + b) $, so we factor the left - hand side: $ (x - 3)(x + 3)=0 $.
## Step 2: Solve for $ x $
If a product of two factors is zero, then at least one of the factors must be zero. So we set each factor equal to zero:
- For $ x - 3 = 0 $, we add 3 to both sides of the equation and get $ x=3 $.
- For $ x + 3 = 0 $, we subtract 3 from both sides of the equation and get $ x=-3 $.

# Answer: $ x = 3 $ or $ x=-3 $

### Problem 20: $ (x + 4)^2=25 $
# Explanation:
## Step 1: Take square roots of both sides
We take the square root of both sides of the equation $ (x + 4)^2 = 25 $. Remember that if $ y^2=k $ (where $ k\geq0 $), then $ y=\pm\sqrt{k} $. So we have $ x + 4=\pm\sqrt{25} $.
## Step 2: Simplify and solve for $ x $
Since $ \sqrt{25}=5 $, the equation becomes $ x + 4=\pm5 $.
- Case 1: When $ x + 4 = 5 $, we subtract 4 from both sides: $ x=5 - 4=1 $.
- Case 2: When $ x + 4=-5 $, we subtract 4 from both sides: $ x=-5 - 4=-9 $.

# Answer: $ x = 1 $ or $ x=-9 $

### Problem 23: $ (x - 7)^2=-8 $
# Explanation:
## Step 1: Analyze the square of a real number
The square of any real number $ a $ is non - negative, i.e., $ a^2\geq0 $ for all real numbers $ a $. In the equation $ (x - 7)^2=-8 $, the left - hand side $ (x - 7)^2 $ is a square of a real number (assuming $ x $ is a real number), and the right - hand side is $ - 8$, which is negative.
## Step 2: Conclusion about the solution
Since a non - negative number cannot be equal to a negative number in the set of real numbers, the equation $ (x - 7)^2=-8 $ has no real solutions. If we consider complex numbers, we can proceed as follows:
Take the square root of both sides: $ x - 7=\pm\sqrt{-8}=\pm2\sqrt{2}i $ (where $ i=\sqrt{-1} $). Then $ x=7\pm2\sqrt{2}i $. But if we are working in the real number system, there is no solution.

# Answer: No real solutions (or $ x = 7\pm2\sqrt{2}i $ in complex numbers)

### Problem 26: $ x^2-6x + 13 = 0 $
# Explanation:
## Step 1: Use the quadratic formula
For a quadratic equation $ ax^2+bx + c = 0 $ ($ a
eq0 $), the quadratic formula is $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $. In the equation $ x^2-6x + 13 = 0 $, we have $ a = 1 $, $ b=-6 $, and $ c = 13 $.
First, calculate the discriminant $ D=b^2-4ac $. Substitute the values: $ D=(-6)^2-4\times1\times13=36 - 52=-16 $.
## Step 2: Solve for $ x $
Since $ D=-16 $, $ \sqrt{D}=\sqrt{-16}=4i $. Then $ x=\frac{-(-6)\pm4i}{2\times1}=\frac{6\pm4i}{2}=3\pm2i $.

# Answer: $ x = 3 + 2i $ or $ x = 3-2i $

### Problem 29: $ 2x^2-x + 30 = 0 $
# Explanation:
## Step 1: Use the quadratic formula
For the quadratic equation $ 2x^2-x + 30 = 0 $, we have $ a = 2 $, $ b=-1 $, and $ c = 30 $.
First, calculate the discriminant $ D=b^2-4ac=(-1)^2-4\times2\times30=1 - 240=-239 $.
## Step 2: Solve for $ x $
Since $ D=-239 $, $ \sqrt{D}=\sqrt{-239}=i\sqrt{239} $. Then $ x=\frac{-(-1)\pm i\sqrt{239}}{2\times2}=\frac{1\pm i\sqrt{239}}{4} $. If we are working in the real number system, there are no real solutions.

# Answer: $ x=\frac{1\pm i\sqrt{239}}{4} $ (or no real solutions)

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QUESTION IMAGE

Question

Explanation:

Problem 17: \( x^2 - 9 = 0 \)

Step 1: Rewrite as difference of squares

Step 2: Solve for \( x \)

Step 1: Take square roots of both sides

Step 2: Simplify and solve for \( x \)

Step 1: Analyze the square of a real number

Step 2: Conclusion about the solution

Answer:

Problem 20: \( (x + 4)^2=25 \)