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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]
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