IDNLearn.com offers a reliable platform for finding accurate and timely answers. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
To solve the equation [tex]\(\sqrt{2 \gamma + 3} + 6 = \gamma\)[/tex], follow these steps:
1. Isolate the square root: First, we want to get the square root term by itself on one side of the equation. Subtract 6 from both sides:
[tex]\[ \sqrt{2 \gamma + 3} = \gamma - 6 \][/tex]
2. Square both sides: To eliminate the square root, square both sides of the equation:
[tex]\[ (\sqrt{2 \gamma + 3})^2 = (\gamma - 6)^2 \][/tex]
This simplifies to:
[tex]\[ 2 \gamma + 3 = (\gamma - 6)^2 \][/tex]
3. Expand the right-hand side: Expand [tex]\((\gamma - 6)^2\)[/tex]:
[tex]\[ 2 \gamma + 3 = \gamma^2 - 12 \gamma + 36 \][/tex]
4. Rearrange into a standard quadratic equation: Move all terms to one side of the equation to form a quadratic equation:
[tex]\[ 2 \gamma + 3 - 2 \gamma = \gamma^2 - 12 \gamma + 36 - 2 \gamma - 3 \][/tex]
Simplify to:
[tex]\[ 0 = \gamma^2 - 14 \gamma + 33 \][/tex]
5. Solve the quadratic equation: Use the quadratic formula, [tex]\(\gamma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = 33\)[/tex]. Plug in these values:
[tex]\[ \gamma = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 33}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ \gamma = \frac{14 \pm sqrt{196 - 132}}{2} \][/tex]
[tex]\[ \gamma = \frac{14 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ \gamma = \frac{14 \pm 8}{2} \][/tex]
This gives us two solutions:
[tex]\[ \gamma = \frac{14 + 8}{2} = \frac{22}{2} = 11 \][/tex]
[tex]\[ \gamma = \frac{14 - 8}{2} = \frac{6}{2} = 3 \][/tex]
6. Verify the solutions: Substitute both [tex]\(\gamma = 11\)[/tex] and [tex]\(\gamma = 3\)[/tex] back into the original equation to ensure they are valid solutions.
For [tex]\(\gamma = 11\)[/tex]:
[tex]\[ \sqrt{2 \cdot 11 + 3} + 6 = \sqrt{22 + 3} + 6 = \sqrt{25} + 6 = 5 + 6 = 11 \][/tex]
This is true, so [tex]\(\gamma = 11\)[/tex] is a valid solution.
For [tex]\(\gamma = 3\)[/tex]:
[tex]\[ \sqrt{2 \cdot 3 + 3} + 6 = \sqrt{6 + 3} + 6 = \sqrt{9} + 6 = 3 + 6 = 9 \][/tex]
This is not true, so [tex]\(\gamma = 3\)[/tex] is not a valid solution.
Therefore, the only solution to the equation [tex]\(\sqrt{2 \gamma + 3} + 6 = \gamma\)[/tex] is:
[tex]\[ \gamma = 11 \][/tex]
1. Isolate the square root: First, we want to get the square root term by itself on one side of the equation. Subtract 6 from both sides:
[tex]\[ \sqrt{2 \gamma + 3} = \gamma - 6 \][/tex]
2. Square both sides: To eliminate the square root, square both sides of the equation:
[tex]\[ (\sqrt{2 \gamma + 3})^2 = (\gamma - 6)^2 \][/tex]
This simplifies to:
[tex]\[ 2 \gamma + 3 = (\gamma - 6)^2 \][/tex]
3. Expand the right-hand side: Expand [tex]\((\gamma - 6)^2\)[/tex]:
[tex]\[ 2 \gamma + 3 = \gamma^2 - 12 \gamma + 36 \][/tex]
4. Rearrange into a standard quadratic equation: Move all terms to one side of the equation to form a quadratic equation:
[tex]\[ 2 \gamma + 3 - 2 \gamma = \gamma^2 - 12 \gamma + 36 - 2 \gamma - 3 \][/tex]
Simplify to:
[tex]\[ 0 = \gamma^2 - 14 \gamma + 33 \][/tex]
5. Solve the quadratic equation: Use the quadratic formula, [tex]\(\gamma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = 33\)[/tex]. Plug in these values:
[tex]\[ \gamma = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 33}}{2 \cdot 1} \][/tex]
Simplify under the square root:
[tex]\[ \gamma = \frac{14 \pm sqrt{196 - 132}}{2} \][/tex]
[tex]\[ \gamma = \frac{14 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ \gamma = \frac{14 \pm 8}{2} \][/tex]
This gives us two solutions:
[tex]\[ \gamma = \frac{14 + 8}{2} = \frac{22}{2} = 11 \][/tex]
[tex]\[ \gamma = \frac{14 - 8}{2} = \frac{6}{2} = 3 \][/tex]
6. Verify the solutions: Substitute both [tex]\(\gamma = 11\)[/tex] and [tex]\(\gamma = 3\)[/tex] back into the original equation to ensure they are valid solutions.
For [tex]\(\gamma = 11\)[/tex]:
[tex]\[ \sqrt{2 \cdot 11 + 3} + 6 = \sqrt{22 + 3} + 6 = \sqrt{25} + 6 = 5 + 6 = 11 \][/tex]
This is true, so [tex]\(\gamma = 11\)[/tex] is a valid solution.
For [tex]\(\gamma = 3\)[/tex]:
[tex]\[ \sqrt{2 \cdot 3 + 3} + 6 = \sqrt{6 + 3} + 6 = \sqrt{9} + 6 = 3 + 6 = 9 \][/tex]
This is not true, so [tex]\(\gamma = 3\)[/tex] is not a valid solution.
Therefore, the only solution to the equation [tex]\(\sqrt{2 \gamma + 3} + 6 = \gamma\)[/tex] is:
[tex]\[ \gamma = 11 \][/tex]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
