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Sagot :
Certainly! Let's solve this step-by-step.
### Part (a): Write \( x^2 - 2x - 6 \) in the form \( (x + c)^2 + d \)
To write the quadratic expression \( x^2 - 2x - 6 \) in the form \( (x+c)^2 + d \), we'll use the method of completing the square. Here we go:
1. Original Expression: \( x^2 - 2x - 6 \)
2. Identify the coefficients: The coefficient of \( x \) is \(-2\), and the constant term is \(-6\).
3. Complete the Square:
- Take the coefficient of \( x \) (which is \(-2\)), divide it by 2, and then square it.
- \(\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1\)
- Add and subtract this square inside the expression:
- \( x^2 - 2x + 1 - 1 - 6 \)
- This can be rewritten as:
- \( (x^2 - 2x + 1) - 1 - 6 \)
- \( (x - 1)^2 - 7 \)
So, \( x^2 - 2x - 6 \) in the form \( (x+c)^2 + d \) is:
[tex]\[ (x - 1)^2 - 7 \][/tex]
Therefore:
[tex]\[ c = -1 \][/tex]
[tex]\[ d = -7 \][/tex]
### Part (b): Solve \( x^2 - 2x - 6 = 0 \) using the completed square form
Using the completed square form \( (x-1)^2 - 7 = 0 \):
1. Set the equation to zero:
[tex]\[ (x - 1)^2 - 7 = 0 \][/tex]
2. Isolate the squared term:
[tex]\[ (x - 1)^2 = 7 \][/tex]
3. Take the square root of both sides:
[tex]\[ x - 1 = \pm \sqrt{7} \][/tex]
4. Solve for \( x \):
[tex]\[ x = 1 \pm \sqrt{7} \][/tex]
In the form \( x = f \pm \sqrt{g} \):
[tex]\[ f = 1 \][/tex]
[tex]\[ g = 7 \][/tex]
### Summary
- The values of \( c \) and \( d \) are \( c = -1 \) and \( d = -7 \).
- The solutions to the equation [tex]\( x^2 - 2x - 6 = 0 \)[/tex] are given by [tex]\( x = 1 \pm \sqrt{7} \)[/tex].
### Part (a): Write \( x^2 - 2x - 6 \) in the form \( (x + c)^2 + d \)
To write the quadratic expression \( x^2 - 2x - 6 \) in the form \( (x+c)^2 + d \), we'll use the method of completing the square. Here we go:
1. Original Expression: \( x^2 - 2x - 6 \)
2. Identify the coefficients: The coefficient of \( x \) is \(-2\), and the constant term is \(-6\).
3. Complete the Square:
- Take the coefficient of \( x \) (which is \(-2\)), divide it by 2, and then square it.
- \(\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1\)
- Add and subtract this square inside the expression:
- \( x^2 - 2x + 1 - 1 - 6 \)
- This can be rewritten as:
- \( (x^2 - 2x + 1) - 1 - 6 \)
- \( (x - 1)^2 - 7 \)
So, \( x^2 - 2x - 6 \) in the form \( (x+c)^2 + d \) is:
[tex]\[ (x - 1)^2 - 7 \][/tex]
Therefore:
[tex]\[ c = -1 \][/tex]
[tex]\[ d = -7 \][/tex]
### Part (b): Solve \( x^2 - 2x - 6 = 0 \) using the completed square form
Using the completed square form \( (x-1)^2 - 7 = 0 \):
1. Set the equation to zero:
[tex]\[ (x - 1)^2 - 7 = 0 \][/tex]
2. Isolate the squared term:
[tex]\[ (x - 1)^2 = 7 \][/tex]
3. Take the square root of both sides:
[tex]\[ x - 1 = \pm \sqrt{7} \][/tex]
4. Solve for \( x \):
[tex]\[ x = 1 \pm \sqrt{7} \][/tex]
In the form \( x = f \pm \sqrt{g} \):
[tex]\[ f = 1 \][/tex]
[tex]\[ g = 7 \][/tex]
### Summary
- The values of \( c \) and \( d \) are \( c = -1 \) and \( d = -7 \).
- The solutions to the equation [tex]\( x^2 - 2x - 6 = 0 \)[/tex] are given by [tex]\( x = 1 \pm \sqrt{7} \)[/tex].
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