Certainly! Let's solve the quadratic equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex] step-by-step.
1. Identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
In the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
- [tex]\( a \)[/tex] is the coefficient of [tex]\( x^2 \)[/tex],
- [tex]\( b \)[/tex] is the coefficient of [tex]\( x \)[/tex],
- [tex]\( c \)[/tex] is the constant term.
For the equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex],
- [tex]\( b = -8 \)[/tex],
- [tex]\( c = 16 \)[/tex].
2. Compute the discriminant [tex]\( \Delta \)[/tex]:
The discriminant for a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = (-8)^2 - 4 \cdot 1 \cdot 16
\][/tex]
[tex]\[
\Delta = 64 - 64
\][/tex]
[tex]\[
\Delta = 0
\][/tex]
3. Find the roots using the quadratic formula:
The roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Since the discriminant [tex]\( \Delta = 0 \)[/tex], the equation has a repeated root. The quadratic formula simplifies to:
[tex]\[
x = \frac{-b}{2a}
\][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = \frac{-(-8)}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{8}{2}
\][/tex]
[tex]\[
x = 4
\][/tex]
Therefore, the equation has a repeated root:
[tex]\[
x_1 = x_2 = 4
\][/tex]
To summarize:
- The coefficients are: [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], [tex]\( c = 16 \)[/tex].
- The discriminant is [tex]\( \Delta = 0 \)[/tex].
- The roots of the equation are [tex]\( x_1 = 4 \)[/tex] and [tex]\( x_2 = 4 \)[/tex].
Thus, the detailed solution for the given quadratic equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex] is:
- [tex]\( a = 1 \)[/tex],
- [tex]\( b = -8 \)[/tex],
- [tex]\( c = 16 \)[/tex],
- Discriminant [tex]\( \Delta = 0 \)[/tex],
- Roots: [tex]\( x_1 = 4.0 \)[/tex] and [tex]\( x_2 = 4.0 \)[/tex].
