Question: Quadratic, Rational, and Radical Equations
Finding the roots of a quadratic equation with leading coefficient 1
Solve for $y$.
y² - 4y + 4 = 0
If there is more than one solution, separate them with commas.
If there is no solution, click on "No solution".
y =
1. **State the problem:** Solve the quadratic equation $$y^2 - 4y + 4 = 0$$ for $y$.
2. **Recall the formula:** For a quadratic equation $$ay^2 + by + c = 0$$, the roots are given by the quadratic formula:
$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Since the leading coefficient $a = 1$, $b = -4$, and $c = 4$.
3. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 4 = 16 - 16 = 0$$
4. **Interpret the discriminant:**
Since $$\Delta = 0$$, there is exactly one real root (a repeated root).
5. **Find the root:**
$$y = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-(-4) \pm \sqrt{0}}{2 \times 1} = \frac{4 \pm 0}{2} = \frac{4}{2} = 2$$
6. **Final answer:**
$$y = 2$$