Choose a quadratic method and derive the formula

Try

Solve x²+8x+7=0 without beginning with the quadratic formula. The first two terms almost form (x+4)²; adding 16 on both sides after moving the constant gives (x+4)²=9. Thus x+4 is 3 or −3, producing x=−1 or x=−7. The two branches come from the equation for a square, not an arbitrary extra symbol.

Predict

Predict which method would expose the structure most quickly for (x−2)(x+5)=0, x²=20 and 3x²+5x−1=0. A correct general formula can solve all three, but it may hide a factorization or square-root relationship already visible. Also predict how a completed square with a negative right-hand side differs over the real and complex numbers.

Experiment

Use this method table as a decision aid, then supply the missing reasoning aloud or in writing.

Initial form	Useful action	Important check
Product equals zero	Set each factor equal to zero	The right side really is zero
Square equals a number	Take both square-root branches	State the number system
General quadratic	Complete a square or use the derived formula	Leading coefficient is nonzero
Leading coefficient zero	Return to a linear equation	Do not divide by zero

Change the worked example’s constant from 7 to 19. Now (x+4)²=−3, so there are no real roots and the complex roots are −4±i√3. Squaring either branch verifies it. A real-coordinate graph has no hidden crossings corresponding to these complex roots.

Understand

Let a,b,c be real with a≠0. Starting from ax²+bx+c=0, divide by a and move the constant to obtain x²+(b/a)x=−c/a. Half the coefficient of x is b/(2a); add its square to both sides. This yields

(x+b/(2a))²=(b²−4ac)/(4a²).

Every step so far is reversible. Define D=b²−4ac. If D is positive, taking both real square-root branches and isolating x gives x=(−b±√D)/(2a). The square root of 4a² is actually 2abs(a). Replacing that denominator by 2a when a is negative exchanges the plus and minus branches, so the same pair results. Forgetting one branch would make that explanation invalid.

If D=0, both branches coincide at −b/(2a), giving a repeated root. If D is negative, write √D=i√(−D) and obtain a conjugate complex pair. If a=0, none of the division by a was legal: solve bx+c=0 directly, including the identity and contradiction cases.

For a fresh numerical completion, 3x²−12x+5=0 becomes x²−4x=−5/3, then (x−2)²=7/3. The roots are 2±√21/3. Substitution into the equivalent form 3(x−2)²−7=0 verifies both exactly. Keeping radicals avoids mistaking rounded near-equality for an exact check.

Apply

The written derivation task requires the chain of reasoning, not merely the final formula. A separate task requires completing a square without quoting that formula. In method-choice work, explain why your chosen representation reduces effort and verify the resulting roots.

When a product equals a nonzero constant, the zero-product rule is unavailable. When a square equals a positive constant, keep both branches. When a real square equals a negative constant, either report no real solutions or explicitly enlarge the number system; do not silently replace the problem by one about complex numbers.

Continue

Quadratic solving becomes especially useful when a rational model is cleared of denominators. The resulting roots are candidates: the original domain and physical meaning still decide which are valid. The next lesson constructs those models rather than starting from a supplied equation.

Continue to Construct rational models from rates.