Solve the following equations:

1. [tex]3t^2 = 12[/tex]

2. [tex]\frac{1}{2}x^2 + 16 = 0[/tex]

3. [tex]x^2 - 7 = 0[/tex]

4. [tex]5x^2 \pm 20 = 5[/tex]

5. [tex]4x^2 + 5 = 21[/tex]

6. [tex]62 \cdot 2x^2 - 128 = 0[/tex]



Answer :

Certainly! Let's solve the given equations step-by-step.

### Equation 1:
[tex]\[ 3t^2 = 12 \][/tex]

1. Divide both sides by 3:
[tex]\[ t^2 = 4 \][/tex]

2. Take the square root of both sides:
[tex]\[ t = \pm 2 \][/tex]

Thus, the solutions are:
[tex]\[ t = -2, \; t = 2 \][/tex]


### Equation 4:
[tex]\[ \frac{1}{2} x^2 + 16 = 0 \][/tex]

1. Subtract 16 from both sides:
[tex]\[ \frac{1}{2} x^2 = -16 \][/tex]

2. Multiply both sides by 2:
[tex]\[ x^2 = -32 \][/tex]

3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{-32} = \pm \sqrt{32} \cdot \sqrt{-1} = \pm 4\sqrt{2}i \][/tex]

Thus, the solutions are:
[tex]\[ x = -4\sqrt{2}i, \; x = 4\sqrt{2}i \][/tex]


### Equation r:
[tex]\[ x^2 - 7 = 0 \][/tex]

1. Add 7 to both sides:
[tex]\[ x^2 = 7 \][/tex]

2. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{7} \][/tex]

Thus, the solutions are:
[tex]\[ x = -\sqrt{7}, \; x = \sqrt{7} \][/tex]


### Equation 5:
[tex]\[ 5x^2 - 20 = 5 \][/tex]

1. Add 20 to both sides:
[tex]\[ 5x^2 = 25 \][/tex]

2. Divide both sides by 5:
[tex]\[ x^2 = 5 \][/tex]

3. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{5} \][/tex]

Thus, the solutions are:
[tex]\[ x = -\sqrt{5}, \; x = \sqrt{5} \][/tex]


### Equation:
[tex]\[ 4x^2 + 5 = 21 \][/tex]

1. Subtract 5 from both sides:
[tex]\[ 4x^2 = 16 \][/tex]

2. Divide both sides by 4:
[tex]\[ x^2 = 4 \][/tex]

3. Take the square root of both sides:
[tex]\[ x = \pm 2 \][/tex]

Thus, the solutions are:
[tex]\[ x = -2, \; x = 2 \][/tex]


### Equation:
[tex]\[ 62 \cdot 2 x^2 - 128 = 0 \][/tex]

1. Distribute the 62:
[tex]\[ 124x^2 - 128 = 0 \][/tex]

2. Add 128 to both sides:
[tex]\[ 124x^2 = 128 \][/tex]

3. Divide both sides by 124:
[tex]\[ x^2 = \frac{128}{124} = \frac{32}{31} \][/tex]

4. Take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{32}{31}} = \pm \frac{\sqrt{32}}{\sqrt{31}} = \pm \frac{4\sqrt{2}}{\sqrt{31}} = \pm \frac{4 \sqrt{62}}{31} \][/tex]

Thus, the solutions are:
[tex]\[ x = -\frac{4 \sqrt{62}}{31}, \; x = \frac{4 \sqrt{62}}{31} \][/tex]


Here is the summary of the solutions:
1. [tex]\( t = -2, \; t = 2 \)[/tex]
4. [tex]\( x = -4\sqrt{2}i, \; x = 4\sqrt{2}i \)[/tex]
r. [tex]\( x = -\sqrt{7}, \; x = \sqrt{7} \)[/tex]
5. [tex]\( x = -\sqrt{5}, \; x = \sqrt{5} \)[/tex]
→ [tex]\( x = -2, \; x = 2 \)[/tex]
[tex]\( x = -\frac{4 \sqrt{62}}{31}, \; x = \frac{4 \sqrt{62}}{31} \)[/tex]