To determine the range of values for [tex]\( k \)[/tex] and find the roots of the equation [tex]\( z^2 - 3z + k = 0 \)[/tex], given that the roots are real numbers, we need to ensure that the quadratic equation has real roots. This involves checking the discriminant of the quadratic equation.
### Quadratic Formula and Discriminant
The general form of a quadratic equation is:
[tex]\[ az^2 + bz + c = 0 \][/tex]
For our specific equation, [tex]\( z^2 - 3z + k = 0 \)[/tex], the coefficients are:
[tex]\[ a = 1, \quad b = -3, \quad c = k \][/tex]
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( az^2 + bz + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the quadratic equation to have real roots, the discriminant must be non-negative:
[tex]\[ \Delta \geq 0 \][/tex]
### Step-by-Step Solution
1. Calculate the Discriminant:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = k \)[/tex]:
[tex]\[
\Delta = (-3)^2 - 4 \cdot 1 \cdot k = 9 - 4k
\][/tex]
2. Ensure Real Roots (Non-negative Discriminant):
[tex]\[
\Delta \geq 0 \Rightarrow 9 - 4k \geq 0
\][/tex]
Solving this inequality:
[tex]\[
9 \geq 4k \quad \Rightarrow \quad k \leq \frac{9}{4} = 2.25
\][/tex]
Therefore, for the quadratic equation [tex]\( z^2 - 3z + k = 0 \)[/tex] to have real roots, [tex]\( k \)[/tex] must satisfy:
[tex]\[ k \leq \frac{9}{4} \][/tex]
3. Find the Roots for a Given Value of [tex]\( k \)[/tex]:
To find the roots, use the quadratic formula:
[tex]\[
z = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( \Delta = 9 - 4k \)[/tex]:
[tex]\[
z = \frac{-(-3) \pm \sqrt{9 - 4k}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 4k}}{2}
\][/tex]
Thus, the roots of the quadratic equation [tex]\( z^2 - 3z + k = 0 \)[/tex] are given by:
[tex]\[ z = \frac{3 + \sqrt{9 - 4k}}{2} \quad \text{and} \quad z = \frac{3 - \sqrt{9 - 4k}}{2} \][/tex]
### Summary
- The value of [tex]\( k \)[/tex] must be [tex]\( k \leq 2.25 \)[/tex] for the roots to be real.
- The roots of the equation [tex]\( z^2 - 3z + k = 0 \)[/tex] are:
[tex]\[
z_1 = \frac{3 + \sqrt{9 - 4k}}{2} \quad \text{and} \quad z_2 = \frac{3 - \sqrt{9 - 4k}}{2}
\][/tex]
