Answer :
To solve the equation [tex]\(\sqrt{x+3} - \sqrt{2x-1} = -2\)[/tex], we can arrange the steps in the following order:
1. Simplify to obtain the final radical term on one side of the equation:
[tex]\[\sqrt{x + 3} = \sqrt{2x - 1} - 2\][/tex]
2. Raise both sides of the equation to the power of 2:
[tex]\[(\sqrt{x + 3})^2 = (\sqrt(2x - 1) - 2)^2\][/tex]
This results in:
[tex]\[x + 3 = (2x - 1) - 4\sqrt{2x - 1} + 4\][/tex]
3. Simplify to get a quadratic equation:
[tex]\[x + 3 = 2x - 1 - 4\sqrt{2x - 1} + 4\][/tex]
[tex]\[x + 3 = 2x + 3 - 4\sqrt{2x - 1}\][/tex]
[tex]\[0 = x - 4\sqrt{2x - 1}\][/tex]
4. Isolate the remaining radical term (though this step isn't numbered, it logically follows the simplification step):
[tex]\[4\sqrt{2x - 1} = x\][/tex]
[tex]\[\sqrt{2x - 1} = \frac{x}{4}\][/tex]
5. Raise both sides of the equation to the power of 2 again:
[tex]\[(\sqrt(2x - 1))^2 = \left(\frac{x}{4}\right)^2\][/tex]
This results in:
[tex]\[2x - 1 = \frac{x^2}{16}\][/tex]
6. Simplify to get the final quadratic equation and use the quadratic formula to find the values of [tex]\(x\)[/tex]:
[tex]\[32x - 16 = x^2\][/tex]
[tex]\[x^2 - 32x + 16 = 0\][/tex]
7. Apply the Zero Product Rule (Use the quadratic formula to solve the quadratic equation):
[tex]\[x = \frac{32 \pm \sqrt{32^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\][/tex]
This results in the solutions:
[tex]\[x_1 \approx 31.491933384829668\][/tex]
[tex]\[x_2 \approx 0.5080666151703319\][/tex]
Upon checking these solutions in the original equation, we find that only [tex]\(x \approx 31.491933384829668\)[/tex] is valid, so the final valid solution is:
[tex]\[x \approx 31.491933384829668\][/tex]
To rearrange the steps as blocks:
- Simplify to obtain the final radical term on one side of the equation.
- Raise both sides of the equation to the power of 2.
- Simplify to get a quadratic equation.
- Raise both sides of the equation to the power of 2 again.
- Use the quadratic formula to find the values of [tex]\(x\)[/tex].
- Apply the Zero Product Rule.
1. Simplify to obtain the final radical term on one side of the equation:
[tex]\[\sqrt{x + 3} = \sqrt{2x - 1} - 2\][/tex]
2. Raise both sides of the equation to the power of 2:
[tex]\[(\sqrt{x + 3})^2 = (\sqrt(2x - 1) - 2)^2\][/tex]
This results in:
[tex]\[x + 3 = (2x - 1) - 4\sqrt{2x - 1} + 4\][/tex]
3. Simplify to get a quadratic equation:
[tex]\[x + 3 = 2x - 1 - 4\sqrt{2x - 1} + 4\][/tex]
[tex]\[x + 3 = 2x + 3 - 4\sqrt{2x - 1}\][/tex]
[tex]\[0 = x - 4\sqrt{2x - 1}\][/tex]
4. Isolate the remaining radical term (though this step isn't numbered, it logically follows the simplification step):
[tex]\[4\sqrt{2x - 1} = x\][/tex]
[tex]\[\sqrt{2x - 1} = \frac{x}{4}\][/tex]
5. Raise both sides of the equation to the power of 2 again:
[tex]\[(\sqrt(2x - 1))^2 = \left(\frac{x}{4}\right)^2\][/tex]
This results in:
[tex]\[2x - 1 = \frac{x^2}{16}\][/tex]
6. Simplify to get the final quadratic equation and use the quadratic formula to find the values of [tex]\(x\)[/tex]:
[tex]\[32x - 16 = x^2\][/tex]
[tex]\[x^2 - 32x + 16 = 0\][/tex]
7. Apply the Zero Product Rule (Use the quadratic formula to solve the quadratic equation):
[tex]\[x = \frac{32 \pm \sqrt{32^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\][/tex]
This results in the solutions:
[tex]\[x_1 \approx 31.491933384829668\][/tex]
[tex]\[x_2 \approx 0.5080666151703319\][/tex]
Upon checking these solutions in the original equation, we find that only [tex]\(x \approx 31.491933384829668\)[/tex] is valid, so the final valid solution is:
[tex]\[x \approx 31.491933384829668\][/tex]
To rearrange the steps as blocks:
- Simplify to obtain the final radical term on one side of the equation.
- Raise both sides of the equation to the power of 2.
- Simplify to get a quadratic equation.
- Raise both sides of the equation to the power of 2 again.
- Use the quadratic formula to find the values of [tex]\(x\)[/tex].
- Apply the Zero Product Rule.