The quadratic formula is short, and it always works, which is exactly why students trust it too quickly. Almost every wrong answer in this chapter comes from the same handful of small slips: a lost minus sign, a square of a negative number done wrongly, a root sign drawn over too little, or a fraction cancelled halfway. The formula itself is rarely the problem.
This guide states the formula, gives a six-step routine to use every time, works through five examples including a word problem, and then lists the slips with their fixes. If you want to know why the formula gives two answers, read why a quadratic equation has two roots first. For the factorisation method that sometimes beats the formula, see the page on solving a quadratic equation by factorisation, and for the whole chapter, the full chapter notes with practice questions.
The Formula and Where It Comes From
For the equation ax² + bx + c = 0 with a ≠ 0,
x = (−b ± √(b² − 4ac)) ÷ 2a
The formula comes from completing the square. Divide by a, move the constant, add (b/2a)² to both sides, and the left side becomes a perfect square, (x + b/2a)² = (b² − 4ac)/4a². Taking the square root of both sides gives the ±, and the formula follows. Knowing this derivation helps you trust each part of the formula, including the 2a that sits under everything.
The quantity b² − 4ac is the discriminant. If it is negative, there are no real roots, and the correct answer is to say so, rather than force a number.
A Six-Step Routine
- Bring the equation to standard form, with everything on one side and zero on the other.
- Write a, b and c with their signs, for example a = 2, b = −7, c = 3.
- Compute D = b² − 4ac separately, with brackets around negative numbers.
- Decide: D > 0 gives two real roots, D = 0 gives equal roots, D < 0 gives none.
- Substitute into the formula, keeping the whole numerator over 2a.
- Simplify, and check by substituting a root back into the original equation, or by testing the sum −b/a and the product c/a.
Five Worked Examples
Example one: a clean case
Solve 2x² − 7x + 3 = 0.
a = 2, b = −7, c = 3. D = (−7)² − 4(2)(3) = 49 − 24 = 25. So x = (7 ± 5) ÷ 4, giving x = 12/4 = 3 or x = 2/4 = 1/2.
Check. Sum = 3 + 1/2 = 7/2, and −b/a = 7/2. Product = 3/2, and c/a = 3/2. ✓
Example two: when the equation is not in standard form
Solve 2x² = 3 − x.
Move everything to one side: 2x² + x − 3 = 0. Now a = 2, b = 1, c = −3. D = 1 − 4(2)(−3) = 1 + 24 = 25. x = (−1 ± 5) ÷ 4, giving 1 or −3/2.
The slip to avoid here is reading the coefficients off the original form, where c looks like +3.
Example three: a root that does not simplify to a whole number
Solve x² − 6x + 6 = 0.
a = 1, b = −6, c = 6. D = 36 − 24 = 12, and √12 = 2√3. So x = (6 ± 2√3) ÷ 2. Divide both terms of the numerator by 2: x = 3 ± √3.
Cancelling only one term, such as writing 6 ± √3, is one of the most common errors in the chapter.
Example four: an equation with a fraction
Solve x + 1/x = 3.
Multiply through by x: x² + 1 = 3x, so x² − 3x + 1 = 0. a = 1, b = −3, c = 1. D = 9 − 4 = 5. x = (3 ± √5) ÷ 2.
Example five: a word problem
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.
Let the speed be x km/h. Then 360/x − 360/(x + 5) = 1. Multiply through by x(x + 5): 360(x + 5) − 360x = x(x + 5), so 1800 = x² + 5x, giving x² + 5x − 1800 = 0.
Here a = 1, b = 5, c = −1800. D = 25 + 7200 = 7225 = 85². So x = (−5 ± 85) ÷ 2, which gives 40 or −45. Speed cannot be negative, so the speed is 40 km/h.
Check. 360 ÷ 40 = 9 hours. At 45 km/h, 360 ÷ 45 = 8 hours, which is 1 hour less. ✓
The Slips and Their Fixes
| Slip | What goes wrong | The fix |
| Wrong sign for b | For b = −7, writing −b as −7 | −b is the opposite of b, so it is +7 |
| Squaring a negative as negative | (−7)² written as −49 | Use brackets: (−7)² = 49 |
| Mishandling 4ac when c is negative | −4(2)(−3) taken as −24 | A negative times a negative is positive, so b² − 4ac becomes b² + 24 |
| Root sign over too little | Taking the root of only b² | The root covers the whole b² − 4ac |
| Dividing only part by 2a | (6 ± 2√3)/2 = 6 ± √3 | Divide every term of the numerator |
| Forgetting the ± | Giving only one root | Always write both |
| Reading a, b, c before standard form | Using coefficients of 2x² = 3 − x directly | Rearrange first |
| Forcing an answer when D < 0 | Writing a root of a negative number | State that there are no real roots |
| Not simplifying the surd | Leaving √12 as it is | Write √12 = 2√3 |
| Never checking | Trusting the arithmetic | Test the sum and product of the roots |
When to Use the Formula and When Not To
The formula always works, but it is not always the fastest. If the equation factorises quickly, as x² − 5x + 6 does into (x − 2)(x − 3), factorisation is shorter. If b = 0, as in x² = 9, take the square root directly. If c = 0, as in x² − 4x = 0, take out the common factor x. The formula is the safe choice when the numbers are awkward, or when factorising by inspection is not obvious. The same “choose the quickest correct tool” habit appears in graphical vs algebraic methods for linear equations.
Quadratic Equations in Your Syllabus
Quadratic Equations is a chapter of the Class 10 Maths book in both the CBSE and the Karnataka SSLC textbooks. Practise with these chapter MCQs, and see the board-style problems collected in the numerical types that appear every year. The formula returns in Class 11, where it is extended to complex roots when D < 0, and it is used constantly in JEE Main and KCET mathematics and physics. Confirm the current portions with your own board.
Frequently Asked Questions
Q1. What is the quadratic formula?
For ax² + bx + c = 0, with a not equal to 0, x = (−b ± √(b² − 4ac)) ÷ 2a.
Q2. What is the most common mistake in using it?
Sign errors, especially forgetting that −b is positive when b is negative, squaring a negative number incorrectly, and mishandling 4ac when c is negative.
Q3. What if the discriminant is negative?
Then there are no real roots, and the correct answer is to say that the equation has no real roots.
Q4. Do I always need the formula?
No. If the equation factorises easily, or if b = 0 or c = 0, a simpler method is quicker.
Q5. How can I check my roots quickly?
Check that the sum of the roots equals −b/a and the product equals c/a, or substitute a root back into the original equation.
Slow Is Smooth, and Smooth Is Fast
The formula punishes haste and rewards a steady routine: brackets around negatives, D worked out on its own line, and every term divided. Those habits cost a few seconds, and they save marks.
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