Choosing a Method, Modeling, and a Line Meeting a Curve

In this lesson you will: Pick the solving method that fits the form of the equation in front of you; Create a quadratic equation from a situation and reject solutions the situation forbids; Solve a system of one linear and one quadratic equation, algebraically and graphically.

Let the form choose the method

Four methods now solve the same equations. They do not all cost the same. A squared quantity sitting alone wants a square root; a friendly trinomial wants factoring; anything else wants the formula.

Completing the square is the one you reach for least often as a solving tool, because the formula already packages it. Its real job is rewriting, which is why Ohio names it separately in A.REI.4a.

From a situation to an equation

A.CED.1b asks you to build the quadratic yourself. The pattern is nearly always the same: name the unknown, write the quantity the problem is about, set it equal to the value given, then solve.

What is new compared with linear modeling is the ending. A quadratic hands back two solutions, and a situation often accepts only one. Lengths are not negative, times before the start do not exist, and a number of people is a whole number.

Rejecting a solution is not crossing out a mistake. Say which one you are discarding and why.

One line, one parabola

A.REI.7 pairs a linear equation with a quadratic one. Solve it by substitution, exactly as in Chapter 12: if both are solved for y, set the two expressions equal.

y=x^2-3 and y=2x give x^2-3=2x, so x^2-2x-3=0 and (x-3)(x+1)=0. Then x=3 or x=-1, and substituting back into the linear equation gives y=6 and y=-2.

The graph shows why two answers are normal. A line can cut a parabola twice, touch it once, or miss it completely — the same three cases the discriminant counts.

Common mistake — Keeping a solution the situation rules out

A path is 6 feet wider than it is long and covers 40 square feet. A student writes w(w+6)=40, solves correctly to get w=4 or w=-10, and reports both as widths.

The algebra really is finished, and both numbers really do satisfy the equation. Nothing in the working looks wrong, because nothing in the working is wrong.

The problem is that the equation is a model, and models describe a narrower world than algebra does. A width of -10 feet is not a width. The model stops applying before the algebra does.

So the answer is 4 feet, with a sentence: -10 is rejected because a width cannot be negative.

How to catch it: After you solve, read each solution back into the first sentence of the problem. Substitute it into the situation, not into the equation: negative lengths, negative times, negative counts and fractional people all fail that reading immediately.