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Theory of quadratic equations: Class 10 Mathematics
Solving quadratics, the discriminant and nature of roots, cube roots of unity, relations between roots and coefficients, symmetric functions, forming equations, and equations that reduce to quadratic form.
Key points
A quadratic equation in x has the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. It can be solved by factoring, by completing the square, or by the quadratic formula x = (−b ± √(b² − 4ac)) / 2a. The formula comes from completing the square on the standard form and works for every quadratic.
The expression D = b² − 4ac is called the discriminant. It is the part under the root sign in the quadratic formula, so it decides what kind of roots the equation has.
Nature of the roots (a, b, c rational):
- If D > 0 and D is a perfect square, the roots are real, rational and unequal.
- If D > 0 and D is not a perfect square, the roots are real, irrational and unequal.
- If D = 0, the roots are real, rational and equal, each equal to −b / 2a.
- If D < 0, the roots are not real. They are a pair of complex conjugates.
Cube roots of unity are the three solutions of x³ = 1. Writing x³ − 1 = (x − 1)(x² + x + 1) = 0 gives x = 1 or x² + x + 1 = 0. The quadratic gives x = (−1 ± i√3) / 2. The two complex roots are named ω and ω², where ω = (−1 + i√3) / 2 and ω² = (−1 − i√3) / 2. Here i² = −1.
Properties of the cube roots of unity:
- 1 + ω + ω² = 0, so the sum of the three roots is zero.
- ω³ = 1, so any power of ω can be reduced by removing multiples of 3 from the index. For example ω⁷ = ω⁶ · ω = ω.
- ω · ω² = 1, so each complex root is the reciprocal of the other.
- 1 + ω = −ω² and 1 + ω² = −ω.
Fourth roots of unity are the solutions of x⁴ = 1. They are 1, −1, i and −i. Their sum is zero and their product is −1.
If α and β are the roots of ax² + bx + c = 0, then the sum of the roots is α + β = −b / a and the product is αβ = c / a. These relations let us find facts about the roots without solving the equation. For example, if the sum is 5 and the product is 3, we can already find α² + β² and 1/α + 1/β.
Symmetric expressions in the roots are rewritten using the sum S and product P:
- α² + β² = S² − 2P.
- α³ + β³ = S³ − 3PS.
- 1/α + 1/β = S / P.
- (α − β)² = S² − 4P.
To form a quadratic equation from its roots α and β, use x² − (α + β)x + αβ = 0. If the roots are given, find their sum and product and put them in this form. If the roots involve a surd, such as 2 + √3 and 2 − √3, the sum and product are rational and the equation has rational coefficients.
Some equations are not quadratic at first but can be turned into quadratics by a substitution. An equation like x⁴ − 5x² + 4 = 0 becomes y² − 5y + 4 = 0 with y = x². An equation with radicals, such as √(x + 6) = x, is solved by squaring both sides, and every answer must be checked in the original equation because squaring can bring in extra roots.
Reciprocal equations like x + 1/x = 5/2 are cleared of fractions first, giving 2x² − 5x + 2 = 0. Exponential equations like 2²ˣ − 5 · 2ˣ + 4 = 0 use the substitution y = 2ˣ. After finding y, go back and solve for x.
Two simultaneous equations, where one is linear and the other is quadratic, are solved by substitution. Find one variable from the linear equation, put it in the quadratic, solve the resulting quadratic, and then find the matching values of the other variable. Each solution is a pair (x, y).
Study tip: Practise the discriminant questions until you can write the nature of roots in one line, as they are easy marks. In questions on the roots, write the sum and product first, then use the symmetric expression formulas. For cube roots of unity, reduce the powers by dividing the index by 3 and always show the step ω³ = 1.
Definitions and flashcards
Say the meaning first, then open the card to check.
Quadratic equation
An equation that can be written as ax² + bx + c = 0 with a ≠ 0.
Root
A value of the variable that satisfies the equation.
Discriminant
The expression b² − 4ac, which tells the nature of the roots.
Cube roots of unity
The three numbers 1, ω and ω² that satisfy x³ = 1.
Complex conjugates
A pair of numbers a + bi and a − bi.
Symmetric function
An expression of the roots that stays the same when the roots are exchanged.
Extraneous root
A value found by squaring that does not satisfy the original equation.
Reciprocal equation
An equation in which replacing x by 1/x gives the same equation.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Quadratic formula | x = (−b ± √(b² − 4ac)) / 2a | Gives both roots of ax² + bx + c = 0. |
| Discriminant | D = b² − 4ac | Its sign decides the nature of the roots. |
| Sum of roots | α + β = −b / a | From the coefficients. |
| Product of roots | αβ = c / a | From the coefficients. |
| Equation from roots | x² − (α + β)x + αβ = 0 | Used to form an equation. |
| Cube roots of unity | 1 + ω + ω² = 0 and ω³ = 1 | Basic properties of ω. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
What is the discriminant of a quadratic equation? State what it tells us. (2 marks)
For ax² + bx + c = 0 the discriminant is D = b² − 4ac. If D > 0 the roots are real and unequal, if D = 0 they are real and equal, and if D < 0 they are complex.
Find the nature of the roots of 3x² + 2x + 5 = 0 without solving it. (2 marks)
Here a = 3, b = 2 and c = 5. D = 4 − 60 = −56, which is negative. So the roots are not real. They are complex conjugates.
State four properties of the cube roots of unity. (3 marks)
The cube roots of unity are 1, ω and ω². First, 1 + ω + ω² = 0. Second, ω³ = 1. Third, ω · ω² = 1. Fourth, 1 + ω = −ω² and 1 + ω² = −ω.
If α and β are the roots of 2x² − 7x + 3 = 0, find α + β and αβ, and then find 1/α + 1/β. (3 marks)
For this equation a = 2, b = −7 and c = 3. The sum is α + β = −b / a = 7/2. The product is αβ = c / a = 3/2. Then 1/α + 1/β = (α + β) / αβ = (7/2) ÷ (3/2) = 7/3.
Show that the roots of x² + x + 1 = 0 are ω and ω². (3 marks)
By the formula, x = (−1 ± √(1 − 4)) / 2 = (−1 ± i√3) / 2. These are ω = (−1 + i√3) / 2 and ω² = (−1 − i√3) / 2. Squaring ω gives (1 − 2i√3 − 3) / 4 = (−2 − 2i√3) / 4 = (−1 − i√3) / 2, which is the other root. So the roots are ω and ω².
Solve the equation x⁴ − 13x² + 36 = 0 and explain the method. (5 marks)
Put y = x². The equation becomes y² − 13y + 36 = 0. This factors as (y − 4)(y − 9) = 0, so y = 4 or y = 9. Since y = x², x² = 4 gives x = ±2, and x² = 9 gives x = ±3. The four solutions are x = 2, −2, 3 and −3. Check x = 2: 16 − 52 + 36 = 0, which is true.
Numericals with worked steps
Find the nature of the roots of 2x² − 5x + 1 = 0.
- a = 2, b = −5, c = 1.
- D = b² − 4ac = 25 − 8 = 17.
- D > 0 and 17 is not a perfect square.
Answer: The roots are real, irrational and unequal.
The roots of x² − 5x + 3 = 0 are α and β. Find α² + β² and α³ + β³.
- Sum S = α + β = 5 and product P = αβ = 3.
- α² + β² = S² − 2P = 25 − 6 = 19.
- α³ + β³ = S³ − 3PS = 125 − 3 × 3 × 5 = 125 − 45 = 80.
Answer: α² + β² = 19 and α³ + β³ = 80.
Form a quadratic equation whose roots are 2 + √3 and 2 − √3.
- Sum = (2 + √3) + (2 − √3) = 4.
- Product = (2 + √3)(2 − √3) = 4 − 3 = 1.
- Equation: x² − (sum)x + product = 0.
Answer: x² − 4x + 1 = 0
Find the values of k for which x² + kx + 9 = 0 has equal roots.
- a = 1, b = k, c = 9.
- For equal roots D = 0, so k² − 36 = 0.
- k² = 36, so k = 6 or k = −6.
Answer: k = 6 or k = −6
Find the value of (1 − ω + ω²)(1 + ω − ω²).
- 1 + ω² = −ω, so 1 − ω + ω² = −ω − ω = −2ω.
- 1 + ω = −ω², so 1 + ω − ω² = −ω² − ω² = −2ω².
- Product = (−2ω)(−2ω²) = 4ω³.
- ω³ = 1, so the product is 4.
Answer: 4
Common mistakes
- Forgetting that a ≠ 0, or putting the wrong signs for b and c into the formula. Write a, b and c on a separate line before substituting.
- Taking the discriminant as √(b² − 4ac) instead of b² − 4ac. The square root is not part of D.
- Using α + β = b / a instead of −b / a. The sum of the roots always has a minus sign.
- Not checking the answers after squaring an equation with a root sign. Extra roots must be rejected.
- Stopping at y after a substitution such as y = x². The final answer must be in terms of the original variable.