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Sagot :
To solve the equation [tex]\( x^2 + 11x + \frac{121}{4} = \frac{125}{4} \)[/tex], let's follow a step-by-step approach to isolate [tex]\( x \)[/tex]:
### Step 1: Simplify and Rearrange the Equation
Given equation:
[tex]\[ x^2 + 11x + \frac{121}{4} = \frac{125}{4} \][/tex]
Subtract [tex]\(\frac{125}{4}\)[/tex] from both sides to set the equation to zero:
[tex]\[ x^2 + 11x + \frac{121}{4} - \frac{125}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x + \frac{121 - 125}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x + \frac{-4}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x - 1 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -1 \)[/tex].
### Step 2: Solve the Quadratic Equation Using the Quadratic Formula
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -1 \)[/tex] into the formula:
[tex]\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-11 \pm \sqrt{121 + 4}}{2} \][/tex]
[tex]\[ x = \frac{-11 \pm \sqrt{125}}{2} \][/tex]
### Step 3: Express [tex]\(\sqrt{125}\)[/tex] in Simpler Terms
[tex]\[ \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \][/tex]
So, the expression becomes:
[tex]\[ x = \frac{-11 \pm 5\sqrt{5}}{2} \][/tex]
### Step 4: Interpret the Results
We now have two possible solutions:
[tex]\[ x_1 = \frac{-11 + 5\sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{-11 - 5\sqrt{5}}{2} \][/tex]
These are the solutions to the quadratic equation.
### Final Values
So, the solutions are:
[tex]\[ x = -\frac{11}{2} + \frac{5\sqrt{5}}{2} \][/tex]
[tex]\[ x = -\frac{11}{2} - \frac{5\sqrt{5}}{2} \][/tex]
Therefore, the solutions for [tex]\( x \)[/tex] in the equation [tex]\( x^2 + 11x + \frac{121}{4} = \frac{125}{4} \)[/tex] are:
[tex]\[ x = -\frac{11}{2} + \frac{5\sqrt{5}}{2} \][/tex]
[tex]\[ x = -\frac{11}{2} - \frac{5\sqrt{5}}{2} \][/tex]
Alternatively written as:
[tex]\[ x = -\frac{11}{2} \pm \frac{5\sqrt{5}}{2} \][/tex]
These solutions match the choices provided in the problem.
### Step 1: Simplify and Rearrange the Equation
Given equation:
[tex]\[ x^2 + 11x + \frac{121}{4} = \frac{125}{4} \][/tex]
Subtract [tex]\(\frac{125}{4}\)[/tex] from both sides to set the equation to zero:
[tex]\[ x^2 + 11x + \frac{121}{4} - \frac{125}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x + \frac{121 - 125}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x + \frac{-4}{4} = 0 \][/tex]
[tex]\[ x^2 + 11x - 1 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -1 \)[/tex].
### Step 2: Solve the Quadratic Equation Using the Quadratic Formula
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -1 \)[/tex] into the formula:
[tex]\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-11 \pm \sqrt{121 + 4}}{2} \][/tex]
[tex]\[ x = \frac{-11 \pm \sqrt{125}}{2} \][/tex]
### Step 3: Express [tex]\(\sqrt{125}\)[/tex] in Simpler Terms
[tex]\[ \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \][/tex]
So, the expression becomes:
[tex]\[ x = \frac{-11 \pm 5\sqrt{5}}{2} \][/tex]
### Step 4: Interpret the Results
We now have two possible solutions:
[tex]\[ x_1 = \frac{-11 + 5\sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{-11 - 5\sqrt{5}}{2} \][/tex]
These are the solutions to the quadratic equation.
### Final Values
So, the solutions are:
[tex]\[ x = -\frac{11}{2} + \frac{5\sqrt{5}}{2} \][/tex]
[tex]\[ x = -\frac{11}{2} - \frac{5\sqrt{5}}{2} \][/tex]
Therefore, the solutions for [tex]\( x \)[/tex] in the equation [tex]\( x^2 + 11x + \frac{121}{4} = \frac{125}{4} \)[/tex] are:
[tex]\[ x = -\frac{11}{2} + \frac{5\sqrt{5}}{2} \][/tex]
[tex]\[ x = -\frac{11}{2} - \frac{5\sqrt{5}}{2} \][/tex]
Alternatively written as:
[tex]\[ x = -\frac{11}{2} \pm \frac{5\sqrt{5}}{2} \][/tex]
These solutions match the choices provided in the problem.
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