Answer :
Sure! Let's solve each of the given equations step by step:
### Equation 1: [tex]\( x^2 = 144 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( x^2 = 144 \)[/tex].
2. Isolate the Variable:
To isolate [tex]\( x \)[/tex], we take the square root of both sides of the equation.
3. Calculate the Roots:
[tex]\[ x = \pm \sqrt{144} \][/tex]
4. Simplify the Square Root:
[tex]\[ \sqrt{144} = 12 \][/tex]
5. Write the Solutions:
[tex]\[ x = \pm 12 \][/tex]
Therefore, the solutions to the equation [tex]\( x^2 = 144 \)[/tex] are:
[tex]\[ x = -12 \quad \text{and} \quad x = 12 \][/tex]
### Equation 2: [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex].
2. Isolate the [tex]\( x^2 \)[/tex] Term:
Add 121 to both sides of the equation:
[tex]\[ 2 \cdot x^2 = 121 \][/tex]
3. Divide Both Sides by 2:
[tex]\[ x^2 = \frac{121}{2} \][/tex]
4. Take the Square Root of Both Sides:
[tex]\[ x = \pm \sqrt{\frac{121}{2}} \][/tex]
5. Simplify the Expression:
[tex]\[ \sqrt{\frac{121}{2}} = \frac{\sqrt{121}}{\sqrt{2}} = \frac{11}{\sqrt{2}} = \frac{11\sqrt{2}}{2} \][/tex]
6. Write the Solutions:
[tex]\[ x = \pm \frac{11\sqrt{2}}{2} \][/tex]
Therefore, the solutions to the equation [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex] are:
[tex]\[ x = -\frac{11\sqrt{2}}{2} \quad \text{and} \quad x = \frac{11\sqrt{2}}{2} \][/tex]
### Equation 3: [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex].
2. Simplify the Coefficient:
Multiply the constants 3 and 5:
[tex]\[ 15 \cdot x^2 - 125 = 0 \][/tex]
3. Isolate the [tex]\( x^2 \)[/tex] Term:
Add 125 to both sides of the equation:
[tex]\[ 15 \cdot x^2 = 125 \][/tex]
4. Divide Both Sides by 15:
[tex]\[ x^2 = \frac{125}{15} = \frac{25}{3} \][/tex]
5. Take the Square Root of Both Sides:
[tex]\[ x = \pm \sqrt{\frac{25}{3}} \][/tex]
6. Simplify the Expression:
[tex]\[ \sqrt{\frac{25}{3}} = \frac{\sqrt{25}}{\sqrt{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \][/tex]
7. Write the Solutions:
[tex]\[ x = \pm \frac{5\sqrt{3}}{3} \][/tex]
Therefore, the solutions to the equation [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex] are:
[tex]\[ x = -\frac{5\sqrt{3}}{3} \quad \text{and} \quad x = \frac{5\sqrt{3}}{3} \][/tex]
### Summary of Solutions
1. For [tex]\( x^2 = 144 \)[/tex]: [tex]\( x = -12 \)[/tex] and [tex]\( x = 12 \)[/tex]
2. For [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex]: [tex]\( x = -\frac{11\sqrt{2}}{2} \)[/tex] and [tex]\( x = \frac{11\sqrt{2}}{2} \)[/tex]
3. For [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex]: [tex]\( x = -\frac{5\sqrt{3}}{3} \)[/tex] and [tex]\( x = \frac{5\sqrt{3}}{3} \)[/tex]
### Equation 1: [tex]\( x^2 = 144 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( x^2 = 144 \)[/tex].
2. Isolate the Variable:
To isolate [tex]\( x \)[/tex], we take the square root of both sides of the equation.
3. Calculate the Roots:
[tex]\[ x = \pm \sqrt{144} \][/tex]
4. Simplify the Square Root:
[tex]\[ \sqrt{144} = 12 \][/tex]
5. Write the Solutions:
[tex]\[ x = \pm 12 \][/tex]
Therefore, the solutions to the equation [tex]\( x^2 = 144 \)[/tex] are:
[tex]\[ x = -12 \quad \text{and} \quad x = 12 \][/tex]
### Equation 2: [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex].
2. Isolate the [tex]\( x^2 \)[/tex] Term:
Add 121 to both sides of the equation:
[tex]\[ 2 \cdot x^2 = 121 \][/tex]
3. Divide Both Sides by 2:
[tex]\[ x^2 = \frac{121}{2} \][/tex]
4. Take the Square Root of Both Sides:
[tex]\[ x = \pm \sqrt{\frac{121}{2}} \][/tex]
5. Simplify the Expression:
[tex]\[ \sqrt{\frac{121}{2}} = \frac{\sqrt{121}}{\sqrt{2}} = \frac{11}{\sqrt{2}} = \frac{11\sqrt{2}}{2} \][/tex]
6. Write the Solutions:
[tex]\[ x = \pm \frac{11\sqrt{2}}{2} \][/tex]
Therefore, the solutions to the equation [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex] are:
[tex]\[ x = -\frac{11\sqrt{2}}{2} \quad \text{and} \quad x = \frac{11\sqrt{2}}{2} \][/tex]
### Equation 3: [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex]
1. Rewrite the Equation:
The original equation is [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex].
2. Simplify the Coefficient:
Multiply the constants 3 and 5:
[tex]\[ 15 \cdot x^2 - 125 = 0 \][/tex]
3. Isolate the [tex]\( x^2 \)[/tex] Term:
Add 125 to both sides of the equation:
[tex]\[ 15 \cdot x^2 = 125 \][/tex]
4. Divide Both Sides by 15:
[tex]\[ x^2 = \frac{125}{15} = \frac{25}{3} \][/tex]
5. Take the Square Root of Both Sides:
[tex]\[ x = \pm \sqrt{\frac{25}{3}} \][/tex]
6. Simplify the Expression:
[tex]\[ \sqrt{\frac{25}{3}} = \frac{\sqrt{25}}{\sqrt{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \][/tex]
7. Write the Solutions:
[tex]\[ x = \pm \frac{5\sqrt{3}}{3} \][/tex]
Therefore, the solutions to the equation [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex] are:
[tex]\[ x = -\frac{5\sqrt{3}}{3} \quad \text{and} \quad x = \frac{5\sqrt{3}}{3} \][/tex]
### Summary of Solutions
1. For [tex]\( x^2 = 144 \)[/tex]: [tex]\( x = -12 \)[/tex] and [tex]\( x = 12 \)[/tex]
2. For [tex]\( 2 \cdot x^2 - 121 = 0 \)[/tex]: [tex]\( x = -\frac{11\sqrt{2}}{2} \)[/tex] and [tex]\( x = \frac{11\sqrt{2}}{2} \)[/tex]
3. For [tex]\( 3 \cdot 5 \cdot x^2 - 125 = 0 \)[/tex]: [tex]\( x = -\frac{5\sqrt{3}}{3} \)[/tex] and [tex]\( x = \frac{5\sqrt{3}}{3} \)[/tex]
