Quadratic formula

For a quadratic equation in standard form $ax^2+bx+c=0$ with $a\neq 0$, the roots are given directly by:

where: $a$, $b$, $c$ are the coefficients of $x^2$, $x$, and the constant term respectively, and the $\pm$ produces the two roots, $x = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $x = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$. The expression under the root, $b^2-4ac$, is called the discriminant, written $D$:

If $D \ge 0$, a real square root exists and both roots above are real numbers. If $D < 0$, no real square root exists, so the equation has no real roots — the full three-way split into $D>0$, $D=0$, and $D<0$ belongs to the next article on the discriminant. The formula itself is not a separate trick to memorise blindly: it is the result of completing the square on $ax^2+bx+c=0$, carried out once so the same steps never have to be repeated for a new equation. The reliable method is: write the equation in standard form, read off $a$, $b$, $c$ carefully (watch every sign), substitute into the formula, simplify the discriminant, then simplify the surd and the fraction. The great advantage of the quadratic formula over factorisation or completing the square is that it works on every quadratic equation, including those whose roots are irrational or involve messy fractions — situations where hunting for a matching pair of integer factors would fail or waste time. This makes it the most dependable method under exam conditions: even when a quick factorisation is visible, applying the formula as a backup check on the answer is good practice, especially when time is short. Because it is derived once from the general equation, it never needs to be re-derived; the discriminant $b^2-4ac$ inside it is not an isolated new quantity to memorise separately but simply the part of the formula that decides whether the square root produces two distinct real numbers, one repeated real number, or no real number at all.

Solve $3x^2 - 5x - 2 = 0$ using the quadratic formula.

Step 1. Identify the coefficients from the standard form $ax^2+bx+c=0$: $a = 3$, $b = -5$, $c = -2$.

Step 2. Compute the discriminant: $b^2 - 4ac = (-5)^2 - 4(3)(-2) = 25 + 24 = 49$. Since $D = 49 > 0$, expect two distinct real roots, and since $49$ is a perfect square, expect them to be rational.

Step 3. Substitute into the formula: $x = \dfrac{-(-5) \pm \sqrt{49}}{2(3)} = \dfrac{5 \pm 7}{6}$.

Step 4. Compute both roots separately: $x = \dfrac{5+7}{6} = \dfrac{12}{6} = 2$ and $x = \dfrac{5-7}{6} = \dfrac{-2}{6} = -\dfrac{1}{3}$. Roots: $x = 2$ or $x = -\dfrac{1}{3}$. Check: $3(2)^2-5(2)-2 = 12-10-2=0$ and $3\left(-\tfrac13\right)^2-5\left(-\tfrac13\right)-2 = \tfrac13+\tfrac53-2 = 0$, confirming both roots.

Competitive stretch

A rectangular park has length 4 m more than its breadth, and area 96 m². Find its dimensions.

Step 1. Let breadth $=x$ m, so length $=(x+4)$ m, and area gives $x(x+4) = 96$, i.e. $x^2+4x-96=0$, with $a=1$, $b=4$, $c=-96$.

Step 2. Discriminant: $b^2-4ac = 16 - 4(1)(-96) = 16+384=400$. Applying the formula: $x = \dfrac{-4 \pm \sqrt{400}}{2} = \dfrac{-4\pm20}{2}$, giving $x=8$ or $x=-12$.

Step 3. Breadth cannot be negative, so reject $x=-12$ and take $x=8$. Breadth $=8$ m, length $=12$ m (check: $8\times12=96$ m²). These two worked cases show the same pattern that appears across the chapter's problem set: a purely numeric equation checks that the mechanics of substitution and surd-simplification are solid, while a mensuration or motion word problem checks that the equation can be set up correctly first, and that the final answer is filtered through common sense about what the unknown represents.

• Sign errors when substituting $b$ and $c$. A negative $b$ or $c$ must be carried carefully into $-b$, $b^2$, and $-4ac$; this is the single most common source of wrong roots.
• Forgetting the $\pm$. Both roots must be written out; stopping after computing one value of $x$ loses a root.
• Simplifying $\sqrt{b^2-4ac}$ incorrectly, such as treating $\sqrt{49}$ as anything other than $7$, or leaving a simplifiable surd like $\sqrt{20}=2\sqrt5$ un-simplified.
• Not rejecting invalid roots in word problems. Lengths, breadths, ages, speeds, and counts of people must be positive; discard the mathematically valid root that fails this physical check.

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