Each of the following quadratic equations can be solved by factoring. Which equation has only one distinct solution?
To solve a quadratic equation by factoring, we want to find two numbers that multiply to give the constant term (the number without a variable) and add to give the coefficient of the linear term (the number multiplied by the variable).
Option 1: $x^2 - 4x + 4$
The constant term is 4 and the coefficient of the linear term is -4. Two numbers that meet these criteria are -2 and -2. Therefore, the equation can be factored as $(x-2)(x-2) = (x-2)^2$. This equation has one solution, $x=2$.
Option 2: $x^2 - 5x + 6$
The constant term is 6 and the coefficient of the linear term is -5. Two numbers that meet these criteria are -2 and -3. Therefore, the equation can be factored as $(x-2)(x-3)$. This equation has two distinct solutions, $x=2$ and $x=3$.
Option 3: $x^2 - 6x + 9$
The constant term is 9 and the coefficient of the linear term is -6. Two numbers that meet these criteria are -3 and -3. Therefore, the equation can be factored as $(x-3)(x-3) = (x-3)^2$. This equation has one solution, $x=3$.
Option 4: $x^2 + 6x + 9$
The constant term is 9 and the coefficient of the linear term is 6. Two numbers that meet these criteria are 3 and 3. Therefore, the equation can be factored as $(x+3)(x+3) = (x+3)^2$. This equation has one solution, $x=-3$.
Of the given options, option 2 ($x^2 - 5x + 6$) is the only one that has two distinct solutions. Therefore, the equation with only one distinct solution is option 1 ($x^2 - 4x + 4$).