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Sagot :
Sure, I can guide you through solving each of these quadratic equations step-by-step.
### Step-by-Step Solutions:
1. Equation [tex]\(5 x^2 - 60 = 0\)[/tex]
- Move the constant term to the right: [tex]\(5 x^2 = 60\)[/tex]
- Divide both sides by 5: [tex]\(x^2 = 12\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{12}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{3}\)[/tex]
2. Equation [tex]\(-7 x^2 + 28 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-7 x^2 = -28\)[/tex]
- Divide both sides by -7: [tex]\(x^2 = 4\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{4}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\)[/tex]
3. Equation [tex]\(-10 x^2 + 50 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-10 x^2 = -50\)[/tex]
- Divide both sides by -10: [tex]\(x^2 = 5\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{5}\)[/tex]
4. Equation [tex]\(6 x^2 - 48 = 0\)[/tex]
- Move the constant term to the right: [tex]\(6 x^2 = 48\)[/tex]
- Divide both sides by 6: [tex]\(x^2 = 8\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{8}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{2}\)[/tex]
5. Equation [tex]\(8 x^2 + 16 = 0\)[/tex]
- Move the constant term to the right: [tex]\(8 x^2 = -16\)[/tex]
- Divide both sides by 8: [tex]\(x^2 = -2\)[/tex]
- Taking the square root of a negative number introduces imaginary numbers: [tex]\(x = \pm i\sqrt{2}\)[/tex]
6. Equation [tex]\(-3 x^2 + 60 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-3 x^2 = -60\)[/tex]
- Divide both sides by -3: [tex]\(x^2 = 20\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{20}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{5}\)[/tex]
7. Equation [tex]\(-5 x^2 - 15 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-5 x^2 = 15\)[/tex]
- Divide both sides by -5: [tex]\(x^2 = -3\)[/tex]
- Taking the square root of a negative number introduces imaginary numbers: [tex]\(x = \pm i\sqrt{3}\)[/tex]
### Summary of Solutions:
1. [tex]\(5 x^2 - 60 = 0 \rightarrow x = \pm 2\sqrt{3}\)[/tex]
2. [tex]\(-7 x^2 + 28 = 0 \rightarrow x = \pm 2\)[/tex]
3. [tex]\(-10 x^2 + 50 = 0 \rightarrow x = \pm \sqrt{5}\)[/tex]
4. [tex]\(6 x^2 - 48 = 0 \rightarrow x = \pm 2\sqrt{2}\)[/tex]
5. [tex]\(8 x^2 + 16 = 0 \rightarrow x = \pm i\sqrt{2}\)[/tex]
6. [tex]\(-3 x^2 + 60 = 0 \rightarrow x = \pm 2\sqrt{5}\)[/tex]
7. [tex]\(-5 x^2 - 15 = 0 \rightarrow x = \pm i\sqrt{3}\)[/tex]
These solutions represent the values of [tex]\(x\)[/tex] that satisfy each of the given quadratic equations.
### Step-by-Step Solutions:
1. Equation [tex]\(5 x^2 - 60 = 0\)[/tex]
- Move the constant term to the right: [tex]\(5 x^2 = 60\)[/tex]
- Divide both sides by 5: [tex]\(x^2 = 12\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{12}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{3}\)[/tex]
2. Equation [tex]\(-7 x^2 + 28 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-7 x^2 = -28\)[/tex]
- Divide both sides by -7: [tex]\(x^2 = 4\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{4}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\)[/tex]
3. Equation [tex]\(-10 x^2 + 50 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-10 x^2 = -50\)[/tex]
- Divide both sides by -10: [tex]\(x^2 = 5\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{5}\)[/tex]
4. Equation [tex]\(6 x^2 - 48 = 0\)[/tex]
- Move the constant term to the right: [tex]\(6 x^2 = 48\)[/tex]
- Divide both sides by 6: [tex]\(x^2 = 8\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{8}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{2}\)[/tex]
5. Equation [tex]\(8 x^2 + 16 = 0\)[/tex]
- Move the constant term to the right: [tex]\(8 x^2 = -16\)[/tex]
- Divide both sides by 8: [tex]\(x^2 = -2\)[/tex]
- Taking the square root of a negative number introduces imaginary numbers: [tex]\(x = \pm i\sqrt{2}\)[/tex]
6. Equation [tex]\(-3 x^2 + 60 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-3 x^2 = -60\)[/tex]
- Divide both sides by -3: [tex]\(x^2 = 20\)[/tex]
- Take the square root of both sides: [tex]\(x = \pm\sqrt{20}\)[/tex]
- Simplify the square root: [tex]\(x = \pm2\sqrt{5}\)[/tex]
7. Equation [tex]\(-5 x^2 - 15 = 0\)[/tex]
- Move the constant term to the right: [tex]\(-5 x^2 = 15\)[/tex]
- Divide both sides by -5: [tex]\(x^2 = -3\)[/tex]
- Taking the square root of a negative number introduces imaginary numbers: [tex]\(x = \pm i\sqrt{3}\)[/tex]
### Summary of Solutions:
1. [tex]\(5 x^2 - 60 = 0 \rightarrow x = \pm 2\sqrt{3}\)[/tex]
2. [tex]\(-7 x^2 + 28 = 0 \rightarrow x = \pm 2\)[/tex]
3. [tex]\(-10 x^2 + 50 = 0 \rightarrow x = \pm \sqrt{5}\)[/tex]
4. [tex]\(6 x^2 - 48 = 0 \rightarrow x = \pm 2\sqrt{2}\)[/tex]
5. [tex]\(8 x^2 + 16 = 0 \rightarrow x = \pm i\sqrt{2}\)[/tex]
6. [tex]\(-3 x^2 + 60 = 0 \rightarrow x = \pm 2\sqrt{5}\)[/tex]
7. [tex]\(-5 x^2 - 15 = 0 \rightarrow x = \pm i\sqrt{3}\)[/tex]
These solutions represent the values of [tex]\(x\)[/tex] that satisfy each of the given quadratic equations.
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