Before you start
Use signed arithmetic, squares, square roots and substitution. Understand that x² means x multiplied by itself. This guide works over real numbers and keeps exact radical answers unless rounding is requested.
Need a starting point? Grade 8 Factorising Expressions: Common Factors and Identities builds a related prerequisite or visual idea.
Put the equation in standard form
A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0. Move all terms to one side and collect like terms before identifying the coefficients. Include each sign. For x² − 7 = 0, a = 1, b = 0 and c = −7.
If a is zero after simplifying, the equation is not quadratic and this formula cannot be used. For instance, 0x² + 3x − 6 = 0 is a linear equation. Do not divide by 2a when that denominator is zero.
Use the formula with brackets
The quadratic formula is x = [−b ± √(b² − 4ac)] ÷ (2a). The whole numerator, not just the square root, is divided by 2a. The plus-or-minus symbol asks you to calculate one result with + and another with −.
Write signed substitutions before calculating. For a = 2, b = −3, c = −4, use −(−3) and (−3)². The square of −3 is 9, while the negative of b is 3. Brackets show which operation applies to which quantity.
Calculate the discriminant first
The expression D = b² − 4ac is the discriminant. If D > 0, there are two distinct real roots. If D = 0, the two formula branches give the same real value, called a repeated root. If D < 0, there are no real roots.
“No real roots” does not mean that a calculation mistake must have occurred. For x² + 1 = 0, every real square is at least zero, so it cannot equal −1. Complex-number solutions belong to a later extension and are outside this guide.
Keep exact answers until rounding is requested
If D is not a perfect square, leave the square root in the answer and simplify when appropriate. For x² − 2 = 0, the exact roots are ±√2. The decimals 1.414 and −1.414 are approximations, not equal replacements for the exact values.
When asked for a decimal answer, keep full calculator precision until the final rounding. Substituting a rounded root into the original may produce a small residual rather than exactly zero. That is different from a large discrepancy caused by a sign error.
Why the formula works
Start from ax² + bx + c = 0 with a ≠ 0. Multiply by 4a and rearrange to get 4a²x² + 4abx = −4ac. Add b² to both sides: (2ax + b)² = b² − 4ac.
Taking both square-root possibilities gives 2ax + b = ±√(b² − 4ac). Subtract b and divide by 2a to obtain the formula. The two signs are needed because a positive square can come from either a positive or negative number.
Choose a method and interpret the answer
Easy factors can be quicker: x² − 5x + 6 = (x − 2)(x − 3). The formula is useful when factoring is not obvious and gives exact real roots whenever the discriminant is non-negative. Completing the square explains the underlying structure.
In a word problem, check units and restrictions after solving. A length must be positive; a count must be a feasible whole number. Reject an inadmissible root with a reason, while retaining both roots when solving an unrestricted algebraic equation.
See the relationships
Read the labels and explanation together. These diagrams illustrate the examples stated in their captions; the practice questions give their own values.
Worked examples, step by step
Example 1: Solve x² − 5x + 6 = 0 using the formula.
- a = 1, b = −5, c = 6. D = 25 − 24 = 1.
- x = (5 ± 1)/2, so x = 3 or x = 2.
- Check: 9 − 15 + 6 = 0 and 4 − 10 + 6 = 0.
Example 2: Solve 2x² − 3x − 4 = 0 exactly.
- a = 2, b = −3, c = −4. D = 9 + 32 = 41.
- x = (3 ± √41)/4. Neither branch should be discarded.
- Check the structure: multiplying a root by 4 gives 3 ± √41. Squaring 4x − 3 gives 41, equivalent to 2x² − 3x − 4 = 0.
Example 3: Solve 3x² − 12 = 0.
- a = 3, b = 0, c = −12. D = 144.
- x = (0 ± 12)/6 = ±2.
- Both roots work because 3(2²) − 12 = 0 and 3((−2)²) − 12 = 0. Directly taking square roots is quicker here.
Example 4: Determine the roots of x² − 6x + 9 = 0.
- D = (−6)² − 4(1)(9) = 0.
- x = (6 ± 0)/2 = 3.
- There is one distinct real value, a repeated root. Check using (x − 3)² = 0.
Example 5: Does x² + 2x + 5 = 0 have real roots?
- D = 2² − 4(1)(5) = −16.
- A negative number has no real square root, so there are no real roots.
- Check by completing the square: (x + 1)² + 4 is always positive for real x.
Example 6: A rectangle has width x cm, length x + 3 cm and area 40 cm². Find its dimensions.
- x(x + 3) = 40 gives x² + 3x − 40 = 0.
- D = 9 + 160 = 169. x = (−3 ± 13)/2 gives 5 or −8.
- Reject −8 cm as a width. The dimensions are 5 cm by 8 cm; their product is 40 cm².
Common mistakes and how to repair them
Using b instead of −b
Write the substitution −(−3) before turning it into +3. The formula uses the negative of the coefficient.
Dividing only the radical by 2a
Use brackets around the entire numerator. Both −b and the signed square-root term share the denominator.
Forgetting the negative root
The ± sign produces two branches unless D = 0. A context may reject one, but an unrestricted equation does not.
Practise before opening the answers
Write your working on paper. Open each answer after making an attempt, then compare the explanation. For construction questions, use the named physical tools.
Identify a, b and c in 3x² + 2x − 1 = 0.
Show answer and reasoning
a = 3, b = 2, c = −1. Keep the sign of the constant.
Find D for x² − 4x + 4 = 0 and state the root type.
Show answer and reasoning
D = 0; repeated real root. 16 − 16 = 0. The root is 2.
Solve x² − 7x + 12 = 0.
Show answer and reasoning
x = 3 or 4. D = 49 − 48 = 1. The roots are (7 ± 1)/2.
Solve 2x² + x − 3 = 0.
Show answer and reasoning
x = 1 or −3/2. D = 1 + 24 = 25. The roots are (−1 ± 5)/4.
Solve x² − 3 = 0 exactly.
Show answer and reasoning
x = ±√3. D = 12, so x = ±√12/2 = ±√3.
Does x² + x + 1 = 0 have real roots?
Show answer and reasoning
No. D = 1 − 4 = −3.
Rearrange x² = 2x + 8 and solve.
Show answer and reasoning
x = 4 or −2. Use x² − 2x − 8 = 0. D = 36, giving (2 ± 6)/2.
A rectangle has width x cm and length x + 1 cm, with area 12 cm². Find both dimensions.
Show answer and reasoning
3 cm and 4 cm. x² + x − 12 = 0 gives x = 3 or −4. Reject the negative width and check 3 × 4 = 12.
Questions students and parents ask
Does the formula work if the x term is missing?
Yes. Use b = 0. For 3x² − 12 = 0 the formula gives ±2, although a direct square-root method is shorter.
Why must a be non-zero?
Otherwise there is no quadratic term and the denominator 2a is zero. Simplify first and solve the resulting linear or constant statement appropriately.
Does D = 0 mean the answer is zero?
No. It means both branches coincide at x = −b/(2a), which may be non-zero. For x² − 6x + 9 = 0, that value is 3.
Should I always turn the roots into decimals?
No. Use exact integers, fractions or radicals unless the question asks for an approximation. Round once at the end when a decimal answer is required.
A simple way to practise together
Ask the learner to write the signed coefficients on a separate line before substitution. Check the discriminant together, then let them calculate both branches and test the answers in the original equation.
Keep learning and practising
- Class 9 Polynomials and Zeros: Meaning, Degree and Examples
- Grade 8 Algebraic Identities with Area Models
For a change of pace, explore our learning games. The games list explains the skills each one practises; not every guide has a matching game level. Return to a fresh question here to check this specific idea.
About this guide
Adapted for independent reading from BananaLearn’s Grade 10 teaching material. We reused our course visual library where it fits and created supporting diagrams for the written explanations. The examples and practice are original learning resources, not official NCERT solutions or reproduced exam questions. School sequences and textbook editions differ.
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