Answer :
Sure, let's go through each equation step-by-step and solve them one by one.
### (1) Solve [tex]\( x^2 = 36 \)[/tex]
To solve the equation [tex]\( x^2 = 36 \)[/tex]:
1. Notice that this is a simple quadratic equation. We can solve for [tex]\( x \)[/tex] by taking the square root of both sides.
[tex]\[ x = \pm \sqrt{36} \][/tex]
2. Since [tex]\( \sqrt{36} = 6 \)[/tex], we get:
[tex]\[ x = \pm 6 \][/tex]
Thus, the solutions are:
[tex]\[ x = -6 \quad \text{and} \quad x = 6 \][/tex]
### (2) Solve [tex]\( (p+8)^2 = 14 \)[/tex]
To solve the equation [tex]\( (p+8)^2 = 14 \)[/tex]:
1. Take the square root of both sides:
[tex]\[ p + 8 = \pm \sqrt{14} \][/tex]
2. Now, solve for [tex]\( p \)[/tex]:
[tex]\[ p = -8 \pm \sqrt{14} \][/tex]
This gives us two solutions:
[tex]\[ p = -8 - \sqrt{14} \quad \text{and} \quad p = -8 + \sqrt{14} \][/tex]
### (3) Solve [tex]\( 8x^2 - 18 = 0 \)[/tex]
To solve the equation [tex]\( 8x^2 - 18 = 0 \)[/tex]:
1. First, isolate the [tex]\( x^2 \)[/tex] term by adding 18 to both sides:
[tex]\[ 8x^2 = 18 \][/tex]
2. Divide both sides by 8:
[tex]\[ x^2 = \frac{18}{8} \][/tex]
[tex]\[ x^2 = \frac{9}{4} \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{9}{4}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{9}}{\sqrt{4}} \][/tex]
[tex]\[ x = \pm \frac{3}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{3}{2} \quad \text{and} \quad x = \frac{3}{2} \][/tex]
### (4) Solve [tex]\( 4x^2 - 1 = 0 \)[/tex]
To solve the equation [tex]\( 4x^2 - 1 = 0 \)[/tex]:
1. Isolate the [tex]\( x^2 \)[/tex] term:
[tex]\[ 4x^2 = 1 \][/tex]
2. Divide both sides by 4:
[tex]\[ x^2 = \frac{1}{4} \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{1}{4}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{1}}{\sqrt{4}} \][/tex]
[tex]\[ x = \pm \frac{1}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{1}{2} \quad \text{and} \quad x = \frac{1}{2} \][/tex]
### (5) Solve [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex]
To solve the equation [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex]:
1. Take the square root of both sides:
[tex]\[ p + 1 = \pm \sqrt{\frac{4}{9}} \][/tex]
[tex]\[ p + 1 = \pm \frac{\sqrt{4}}{\sqrt{9}} \][/tex]
[tex]\[ p + 1 = \pm \frac{2}{3} \][/tex]
2. Now solve for [tex]\( p \)[/tex]:
[tex]\[ p = -1 \pm \frac{2}{3} \][/tex]
This gives us two solutions:
[tex]\[ p = -1 + \frac{2}{3} = -\frac{1}{3} \][/tex]
[tex]\[ p = -1 - \frac{2}{3} = -\frac{5}{3} \][/tex]
Thus, the solutions are:
[tex]\[ p = -\frac{5}{3} \quad \text{and} \quad p = -\frac{1}{3} \][/tex]
So to summarize, the solutions are:
1. [tex]\( x^2 = 36 \)[/tex] ⟹ [tex]\( x = -6, 6 \)[/tex]
2. [tex]\( (p+8)^2 = 14 \)[/tex] ⟹ [tex]\( p = -8 - \sqrt{14}, -8 + \sqrt{14} \)[/tex]
3. [tex]\( 8x^2 - 18 = 0 \)[/tex] ⟹ [tex]\( x = -\frac{3}{2}, \frac{3}{2} \)[/tex]
4. [tex]\( 4x^2 - 1 = 0 \)[/tex] ⟹ [tex]\( x = -\frac{1}{2}, \frac{1}{2} \)[/tex]
5. [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex] ⟹ [tex]\( p = -\frac{5}{3}, -\frac{1}{3} \)[/tex]
### (1) Solve [tex]\( x^2 = 36 \)[/tex]
To solve the equation [tex]\( x^2 = 36 \)[/tex]:
1. Notice that this is a simple quadratic equation. We can solve for [tex]\( x \)[/tex] by taking the square root of both sides.
[tex]\[ x = \pm \sqrt{36} \][/tex]
2. Since [tex]\( \sqrt{36} = 6 \)[/tex], we get:
[tex]\[ x = \pm 6 \][/tex]
Thus, the solutions are:
[tex]\[ x = -6 \quad \text{and} \quad x = 6 \][/tex]
### (2) Solve [tex]\( (p+8)^2 = 14 \)[/tex]
To solve the equation [tex]\( (p+8)^2 = 14 \)[/tex]:
1. Take the square root of both sides:
[tex]\[ p + 8 = \pm \sqrt{14} \][/tex]
2. Now, solve for [tex]\( p \)[/tex]:
[tex]\[ p = -8 \pm \sqrt{14} \][/tex]
This gives us two solutions:
[tex]\[ p = -8 - \sqrt{14} \quad \text{and} \quad p = -8 + \sqrt{14} \][/tex]
### (3) Solve [tex]\( 8x^2 - 18 = 0 \)[/tex]
To solve the equation [tex]\( 8x^2 - 18 = 0 \)[/tex]:
1. First, isolate the [tex]\( x^2 \)[/tex] term by adding 18 to both sides:
[tex]\[ 8x^2 = 18 \][/tex]
2. Divide both sides by 8:
[tex]\[ x^2 = \frac{18}{8} \][/tex]
[tex]\[ x^2 = \frac{9}{4} \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{9}{4}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{9}}{\sqrt{4}} \][/tex]
[tex]\[ x = \pm \frac{3}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{3}{2} \quad \text{and} \quad x = \frac{3}{2} \][/tex]
### (4) Solve [tex]\( 4x^2 - 1 = 0 \)[/tex]
To solve the equation [tex]\( 4x^2 - 1 = 0 \)[/tex]:
1. Isolate the [tex]\( x^2 \)[/tex] term:
[tex]\[ 4x^2 = 1 \][/tex]
2. Divide both sides by 4:
[tex]\[ x^2 = \frac{1}{4} \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{1}{4}} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{1}}{\sqrt{4}} \][/tex]
[tex]\[ x = \pm \frac{1}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{1}{2} \quad \text{and} \quad x = \frac{1}{2} \][/tex]
### (5) Solve [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex]
To solve the equation [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex]:
1. Take the square root of both sides:
[tex]\[ p + 1 = \pm \sqrt{\frac{4}{9}} \][/tex]
[tex]\[ p + 1 = \pm \frac{\sqrt{4}}{\sqrt{9}} \][/tex]
[tex]\[ p + 1 = \pm \frac{2}{3} \][/tex]
2. Now solve for [tex]\( p \)[/tex]:
[tex]\[ p = -1 \pm \frac{2}{3} \][/tex]
This gives us two solutions:
[tex]\[ p = -1 + \frac{2}{3} = -\frac{1}{3} \][/tex]
[tex]\[ p = -1 - \frac{2}{3} = -\frac{5}{3} \][/tex]
Thus, the solutions are:
[tex]\[ p = -\frac{5}{3} \quad \text{and} \quad p = -\frac{1}{3} \][/tex]
So to summarize, the solutions are:
1. [tex]\( x^2 = 36 \)[/tex] ⟹ [tex]\( x = -6, 6 \)[/tex]
2. [tex]\( (p+8)^2 = 14 \)[/tex] ⟹ [tex]\( p = -8 - \sqrt{14}, -8 + \sqrt{14} \)[/tex]
3. [tex]\( 8x^2 - 18 = 0 \)[/tex] ⟹ [tex]\( x = -\frac{3}{2}, \frac{3}{2} \)[/tex]
4. [tex]\( 4x^2 - 1 = 0 \)[/tex] ⟹ [tex]\( x = -\frac{1}{2}, \frac{1}{2} \)[/tex]
5. [tex]\( (p+1)^2 = \frac{4}{9} \)[/tex] ⟹ [tex]\( p = -\frac{5}{3}, -\frac{1}{3} \)[/tex]
