Question

each of the following quadratic equations can be solved by factoring. which equation has only one distinct solution? (1 point)
\\( x^2 - 4 = 0 \\)
\\( x^2 - 5x + 4 = 0 \\)
\\( x^2 + 5x + 4 = 0 \\)
\\( x^2 + 4x + 4 = 0 \\)

Explanation:

Step1: Recall the discriminant or factoring for distinct solutions

A quadratic equation \(ax^2 + bx + c = 0\) has a repeated (one distinct) solution when it is a perfect square trinomial, i.e., it factors into \((x + m)^2 = 0\) (or \((x - m)^2 = 0\)), giving a single solution with multiplicity two. We can also use the discriminant \(D = b^2 - 4ac\); if \(D = 0\), there is one distinct solution.

Step2: Analyze each option

• Option 1: \(x^2 - 4 = 0\)

Factor as \((x - 2)(x + 2)=0\). Solutions: \(x = 2\) and \(x=-2\) (two distinct solutions).

• Option 2: \(x^2 - 5x + 4 = 0\)

Factor as \((x - 1)(x - 4)=0\). Solutions: \(x = 1\) and \(x = 4\) (two distinct solutions).

• Option 3: \(x^2 + 5x + 4 = 0\)

Factor as \((x + 1)(x + 4)=0\). Solutions: \(x=-1\) and \(x=-4\) (two distinct solutions).

• Option 4: \(x^2 + 4x + 4 = 0\)

Factor as \((x + 2)^2 = 0\). Solution: \(x=-2\) (repeated, so one distinct solution).

Answer:

\(x^2 + 4x + 4 = 0\) (the fourth option, or D. \(x^2 + 4x + 4 = 0\) if we consider the options as labeled, though the original labels are circles; the equation is \(x^2 + 4x + 4 = 0\))