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Sagot :
Sure! Let's solve the radical equation step-by-step to determine the correct statement about its solutions.
### Given Equation
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
### Step 1: Isolate the Square Root
First, we need to isolate the square root term if it's not already isolated:
[tex]\[ \sqrt{-4x + 28} = x - 7 \][/tex]
### Step 2: Square Both Sides
Next, to eliminate the square root, we square both sides of the equation:
[tex]\[ (\sqrt{-4x + 28})^2 = (x - 7)^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ -4x + 28 = x^2 - 14x + 49 \][/tex]
### Step 3: Rearrange and Form a Quadratic Equation
Now, we'll move all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 14x + 49 + 4x - 28 = 0 \][/tex]
[tex]\[ x^2 - 10x + 21 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Now, we solve the quadratic equation [tex]\( x^2 - 10x + 21 = 0 \)[/tex] using the quadratic formula [tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]. For this equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 21 \)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-10)^2 - 4(1)(21) \][/tex]
[tex]\[ \Delta = 100 - 84 \][/tex]
[tex]\[ \Delta = 16 \][/tex]
Since the discriminant is positive, the quadratic equation has two solutions:
[tex]\[ x = \frac{10 \pm \sqrt{16}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{10 \pm 4}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10 + 4}{2} = 7 \][/tex]
[tex]\[ x = \frac{10 - 4}{2} = 3 \][/tex]
### Step 5: Check for Extraneous Solutions
It's important to check these solutions in the original equation to see if they are valid.
For [tex]\( x = 7 \)[/tex]:
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
[tex]\[ 7 - 7 = \sqrt{-4(7) + 28} \][/tex]
[tex]\[ 0 = \sqrt{-28 + 28} \][/tex]
[tex]\[ 0 = \sqrt{0} \][/tex]
[tex]\[ 0 = 0 \][/tex]
This is true.
For [tex]\( x = 3 \)[/tex]:
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
[tex]\[ 3 - 7 = \sqrt{-4(3) + 28} \][/tex]
[tex]\[ -4 = \sqrt{-12 + 28} \][/tex]
[tex]\[ -4 = \sqrt{16} \][/tex]
[tex]\[ -4 = 4 \][/tex]
This is false.
Therefore, [tex]\( x = 3 \)[/tex] is an extraneous solution.
### Conclusion
The equation has one true solution, which is [tex]\( x = 7 \)[/tex].
So, the correct statement is:
"There is one true solution, with a value greater than 6."
### Given Equation
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
### Step 1: Isolate the Square Root
First, we need to isolate the square root term if it's not already isolated:
[tex]\[ \sqrt{-4x + 28} = x - 7 \][/tex]
### Step 2: Square Both Sides
Next, to eliminate the square root, we square both sides of the equation:
[tex]\[ (\sqrt{-4x + 28})^2 = (x - 7)^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ -4x + 28 = x^2 - 14x + 49 \][/tex]
### Step 3: Rearrange and Form a Quadratic Equation
Now, we'll move all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 14x + 49 + 4x - 28 = 0 \][/tex]
[tex]\[ x^2 - 10x + 21 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Now, we solve the quadratic equation [tex]\( x^2 - 10x + 21 = 0 \)[/tex] using the quadratic formula [tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]. For this equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 21 \)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-10)^2 - 4(1)(21) \][/tex]
[tex]\[ \Delta = 100 - 84 \][/tex]
[tex]\[ \Delta = 16 \][/tex]
Since the discriminant is positive, the quadratic equation has two solutions:
[tex]\[ x = \frac{10 \pm \sqrt{16}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{10 \pm 4}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10 + 4}{2} = 7 \][/tex]
[tex]\[ x = \frac{10 - 4}{2} = 3 \][/tex]
### Step 5: Check for Extraneous Solutions
It's important to check these solutions in the original equation to see if they are valid.
For [tex]\( x = 7 \)[/tex]:
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
[tex]\[ 7 - 7 = \sqrt{-4(7) + 28} \][/tex]
[tex]\[ 0 = \sqrt{-28 + 28} \][/tex]
[tex]\[ 0 = \sqrt{0} \][/tex]
[tex]\[ 0 = 0 \][/tex]
This is true.
For [tex]\( x = 3 \)[/tex]:
[tex]\[ x - 7 = \sqrt{-4x + 28} \][/tex]
[tex]\[ 3 - 7 = \sqrt{-4(3) + 28} \][/tex]
[tex]\[ -4 = \sqrt{-12 + 28} \][/tex]
[tex]\[ -4 = \sqrt{16} \][/tex]
[tex]\[ -4 = 4 \][/tex]
This is false.
Therefore, [tex]\( x = 3 \)[/tex] is an extraneous solution.
### Conclusion
The equation has one true solution, which is [tex]\( x = 7 \)[/tex].
So, the correct statement is:
"There is one true solution, with a value greater than 6."
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