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Sagot :
To solve the equation [tex]\( x^2 - 11 = -4x \)[/tex] by completing the square, we'll follow a systematic approach. Here's the detailed step-by-step solution:
1. Rewrite the equation in the standard quadratic form: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ x^2 - 11 + 4x = 0 \][/tex]
Simplify it to standard form:
[tex]\[ x^2 + 4x - 11 = 0 \][/tex]
2. Isolate the quadratic and linear terms: We want to complete the square, so we keep the [tex]\( x \)[/tex]-terms on one side and the constant term on the other side.
[tex]\[ x^2 + 4x = 11 \][/tex]
3. Complete the square:
- Take the coefficient of [tex]\( x \)[/tex], which is 4.
- Divide it by 2: [tex]\( \frac{4}{2} = 2 \)[/tex].
- Square the result: [tex]\( 2^2 = 4 \)[/tex].
- Add this square to both sides of the equation to maintain equality:
[tex]\[ x^2 + 4x + 4 = 11 + 4 \][/tex]
4. Rewrite the left side as a perfect square trinomial:
The left side of the equation can now be expressed as the square of a binomial:
[tex]\[ (x + 2)^2 = 15 \][/tex]
5. Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x + 2 = \pm \sqrt{15} \][/tex]
6. Isolate [tex]\( x \)[/tex] by solving both resulting equations:
[tex]\[ x + 2 = \sqrt{15} \quad \text{and} \quad x + 2 = -\sqrt{15} \][/tex]
Solving these two equations:
- For [tex]\( x + 2 = \sqrt{15} \)[/tex]:
[tex]\[ x = \sqrt{15} - 2 \][/tex]
- For [tex]\( x + 2 = -\sqrt{15} \)[/tex]:
[tex]\[ x = -\sqrt{15} - 2 \][/tex]
The solutions to the equation are:
[tex]\[ x = \sqrt{15} - 2 \quad \text{and} \quad x = -\sqrt{15} - 2 \][/tex]
Approximating the values:
[tex]\[ x \approx 1.873 \quad \text{and} \quad x \approx -5.873 \][/tex]
Thus, the solutions to the equation [tex]\( x^2 - 11 = -4x \)[/tex] are approximately:
[tex]\[ x \approx 1.873 \quad \text{and} \quad x \approx -5.873 \][/tex]
1. Rewrite the equation in the standard quadratic form: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ x^2 - 11 + 4x = 0 \][/tex]
Simplify it to standard form:
[tex]\[ x^2 + 4x - 11 = 0 \][/tex]
2. Isolate the quadratic and linear terms: We want to complete the square, so we keep the [tex]\( x \)[/tex]-terms on one side and the constant term on the other side.
[tex]\[ x^2 + 4x = 11 \][/tex]
3. Complete the square:
- Take the coefficient of [tex]\( x \)[/tex], which is 4.
- Divide it by 2: [tex]\( \frac{4}{2} = 2 \)[/tex].
- Square the result: [tex]\( 2^2 = 4 \)[/tex].
- Add this square to both sides of the equation to maintain equality:
[tex]\[ x^2 + 4x + 4 = 11 + 4 \][/tex]
4. Rewrite the left side as a perfect square trinomial:
The left side of the equation can now be expressed as the square of a binomial:
[tex]\[ (x + 2)^2 = 15 \][/tex]
5. Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x + 2 = \pm \sqrt{15} \][/tex]
6. Isolate [tex]\( x \)[/tex] by solving both resulting equations:
[tex]\[ x + 2 = \sqrt{15} \quad \text{and} \quad x + 2 = -\sqrt{15} \][/tex]
Solving these two equations:
- For [tex]\( x + 2 = \sqrt{15} \)[/tex]:
[tex]\[ x = \sqrt{15} - 2 \][/tex]
- For [tex]\( x + 2 = -\sqrt{15} \)[/tex]:
[tex]\[ x = -\sqrt{15} - 2 \][/tex]
The solutions to the equation are:
[tex]\[ x = \sqrt{15} - 2 \quad \text{and} \quad x = -\sqrt{15} - 2 \][/tex]
Approximating the values:
[tex]\[ x \approx 1.873 \quad \text{and} \quad x \approx -5.873 \][/tex]
Thus, the solutions to the equation [tex]\( x^2 - 11 = -4x \)[/tex] are approximately:
[tex]\[ x \approx 1.873 \quad \text{and} \quad x \approx -5.873 \][/tex]
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