Question

#4 x² - 2x + 1; a = 1, b = 2, c = 1; (x - 2)(x + 4); and some hand - written division - like expressions

Explanation:

Step1: Identify the quadratic formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For the equation \(x^2-2x + 1 = 0\), we have \(a = 1\), \(b=-2\), \(c = 1\).

Step2: Calculate the discriminant \(\Delta=b^2-4ac\)

Substitute \(a = 1\), \(b=-2\), \(c = 1\) into the discriminant formula: \(\Delta=(-2)^2-4\times1\times1\)
\(=4 - 4=0\)

Step3: Apply the quadratic formula

Since \(\Delta = 0\), we have \(x=\frac{-(-2)\pm\sqrt{0}}{2\times1}=\frac{2\pm0}{2}\)

Step4: Solve for \(x\)

\(x=\frac{2 + 0}{2}=1\) and \(x=\frac{2-0}{2}=1\) (a repeated root)

Alternatively, we can factor the quadratic:

Step1: Recognize the perfect square trinomial

The quadratic \(x^2-2x + 1\) is a perfect square trinomial of the form \(a^2-2ab + b^2=(a - b)^2\). Here, \(a=x\) and \(b = 1\), so \(x^2-2x + 1=(x - 1)^2\)

Step2: Solve \((x - 1)^2=0\)

Taking the square root of both sides, we get \(x-1=0\), so \(x = 1\) (repeated root)

The factoring attempt in the image \((x - 2)(x + 4)=x^2+2x-8\) which is not equal to \(x^2-2x + 1\), so that factoring is incorrect.

Answer:

The correct solution for \(x^2-2x + 1 = 0\) is \(x = 1\) (with multiplicity 2). The factoring \((x - 2)(x + 4)\) is incorrect; the correct factoring is \((x - 1)^2\).