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A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The presence of the \( x^2 \) term is what makes it a quadratic. This is a fundamental concept in algebra, as it's one of the simplest types of equations that involve powers greater than one. Quadratic equations can model various real-world situations, such as projectile motion in physics or calculating areas in geometry.

Understanding the structure of a quadratic equation is crucial as it helps in identifying the components necessary for solving it. For example, in the expression \( 2y^2 - 4y + 2 \), \( a = 2 \), \( b = -4 \), and \( c = 2 \). Recognizing these components is the first step towards factorization, finding the roots, or even graphing the equation.

The roots of a quadratic equation are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). There can be zero, one, or two real roots, depending on the discriminant \( b^2 - 4ac \).

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root, also known as a repeated root.
• If \( b^2 - 4ac < 0 \), there are no real roots, but two complex roots.

In the problem \( 2y^2 - 4y + 2 = 0 \), calculating the discriminant gives \( (-4)^2 - 4 \times 2 \times 2 = 0 \). This indicates there is exactly one real root. However, due to the symmetry of quadratic equations and the form \( (y - r_1)^2 \), this implies both roots are actually the same, \( y = 1 \), which is why the equation can be factored into \( (2y - 1)(y - 1) \).

The quadratic formula is a reliable method to find the roots of any quadratic equation. It is given as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula works by allowing you to input the values of \( a \), \( b \), and \( c \) from your quadratic equation \( ax^2 + bx + c = 0 \). The "plus-minus" sign indicates the two possible roots you can obtain by evaluating both the plus and the minus result.

For the equation \( 2y^2 - 4y + 2 = 0 \), if you solve using the quadratic formula, substitute \( a = 2 \), \( b = -4 \), and \( c = 2 \) into the formula:

\[y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 2 \times 2}}{2 \times 2}\]

Simplifying further:\
\[y = \frac{4 \pm \sqrt{0}}{4} = 1\]

This confirms the roots\( y = 1 \). Hence, the quadratic expression can again be factored into the form \( (2y - 1)(y - 1) \). This technique ensures clarity and precision, particularly for more complex quadratics where factoring by inspection is impractical.