Quadratic formula and models

Identify coefficients, retain both possible roots and reject values only for a stated domain reason.

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Put the equation in a standard form

For ax² + bx + c = 0 with a ≠ 0, the quadratic formula is

x = (-b± square root of (b²-4ac))/(2a).

The denominator is the entire 2a and the square root covers the entire discriminant b²-4ac. The symbol ± requests both the plus and minus calculations.

Solve 2x² + x-3 = 0: a = 2, b = 1, c = -3. The discriminant is 1-4(2)(-3) = 25. Thus x = (-1±5)/4, giving x = 1 or x = -1.5. Substitute each in the original to check it.

For x²-3x-1 = 0, use b = -3, so -b = 3. The roots are (3± square root of 13)/2, approximately 3.303 and −0.303. Keep exact values or extra calculator digits until the requested final rounding.

The discriminant controls real roots

If b²-4ac is positive, there are two distinct real roots. Zero gives one repeated real root. A negative value has no real square root, so there are no real solutions. This statement is about the real-number domain.

A model can exclude one mathematical root

A rectangle has width w cm and length (w + 2) cm, with area 24 cm². The model is w(w + 2) = 24, or w² + 2w-24 = 0. The formula gives w = 4 or w = -6. A physical width must be positive, so use 4 cm and length 6 cm. The negative value is a root of the equation, but not a valid dimension.

Check your understanding 1 The sign is part of the coefficient

Solve x²-4x-2 = 0 exactly, then to two decimal places.

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a = 1, b = -4, c = -2. The roots are (4± square root of 24)/2 = 2± square root of 6, approximately 4.45 and −0.45. Both real roots are retained because no further domain restriction is stated.

Do not replace a model condition with a guess

If a calculated time is −2 s or 5 s and the model is defined only for t ≥ 0, which result applies?

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Only 5 s lies in the modelled interval. The rejection follows from the stated time origin and domain, not from a rule that every negative algebraic answer is wrong.

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