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Sagot :
Sure, let’s complete the square for the equation [tex]\( 0 = x^2 + x - 20 \)[/tex].
1. Move the constant term to the other side:
[tex]\[ x^2 + x = 20 \][/tex]
2. To complete the square, take half of the coefficient of [tex]\( x \)[/tex], square it, and add it to both sides of the equation.
- The coefficient of [tex]\( x \)[/tex] is 1.
- Half of it is [tex]\( \frac{1}{2} \)[/tex].
- Squaring it gives [tex]\( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex].
3. Add [tex]\( \frac{1}{4} \)[/tex] to both sides:
[tex]\[ x^2 + x + \frac{1}{4} = 20 + \frac{1}{4} \][/tex]
4. Now the left side is a perfect square trinomial and can be written as:
[tex]\[ \left(x + \frac{1}{2}\right)^2 = 20.25 \][/tex]
5. Take the square root of both sides:
[tex]\[ x + \frac{1}{2} = \pm \sqrt{20.25} \][/tex]
6. Simplify the square root:
[tex]\[ x + \frac{1}{2} = \pm 4.5 \][/tex]
7. Isolate [tex]\( x \)[/tex] by subtracting [tex]\( \frac{1}{2} \)[/tex] from both sides:
[tex]\[ x = -\frac{1}{2} \pm 4.5 \][/tex]
8. Solve for both possible values of [tex]\( x \)[/tex]:
- [tex]\( x_1 = -\frac{1}{2} + 4.5 = 4.0 \)[/tex]
- [tex]\( x_2 = -\frac{1}{2} - 4.5 = -5.0 \)[/tex]
Therefore, [tex]\( x = 4.0 \)[/tex] and [tex]\( x = -5.0 \)[/tex].
Putting the equation in the completed-square form [tex]\(20 + \square = x^2 + x + \square\)[/tex], your answers are:
[tex]\[ \begin{array}{l} 20 + \frac{1}{4} = x^2 + x + \frac{1}{4} \\ x = -5 \\ x = 4 \end{array} \][/tex]
1. Move the constant term to the other side:
[tex]\[ x^2 + x = 20 \][/tex]
2. To complete the square, take half of the coefficient of [tex]\( x \)[/tex], square it, and add it to both sides of the equation.
- The coefficient of [tex]\( x \)[/tex] is 1.
- Half of it is [tex]\( \frac{1}{2} \)[/tex].
- Squaring it gives [tex]\( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex].
3. Add [tex]\( \frac{1}{4} \)[/tex] to both sides:
[tex]\[ x^2 + x + \frac{1}{4} = 20 + \frac{1}{4} \][/tex]
4. Now the left side is a perfect square trinomial and can be written as:
[tex]\[ \left(x + \frac{1}{2}\right)^2 = 20.25 \][/tex]
5. Take the square root of both sides:
[tex]\[ x + \frac{1}{2} = \pm \sqrt{20.25} \][/tex]
6. Simplify the square root:
[tex]\[ x + \frac{1}{2} = \pm 4.5 \][/tex]
7. Isolate [tex]\( x \)[/tex] by subtracting [tex]\( \frac{1}{2} \)[/tex] from both sides:
[tex]\[ x = -\frac{1}{2} \pm 4.5 \][/tex]
8. Solve for both possible values of [tex]\( x \)[/tex]:
- [tex]\( x_1 = -\frac{1}{2} + 4.5 = 4.0 \)[/tex]
- [tex]\( x_2 = -\frac{1}{2} - 4.5 = -5.0 \)[/tex]
Therefore, [tex]\( x = 4.0 \)[/tex] and [tex]\( x = -5.0 \)[/tex].
Putting the equation in the completed-square form [tex]\(20 + \square = x^2 + x + \square\)[/tex], your answers are:
[tex]\[ \begin{array}{l} 20 + \frac{1}{4} = x^2 + x + \frac{1}{4} \\ x = -5 \\ x = 4 \end{array} \][/tex]
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