Quadratic Equations ISC Class 11: Roots, Nature and Graphs

What Quadratic Equations is about

Weightage: Quadratic Equations belongs to the unit Algebra, which CISCE weights at 26 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

The quadratic formula and complex roots

Dividing ax² + bx + c = 0 by a and completing the square: (x + b/(2a))² = (b² − 4ac)/(4a²), so x = (−b ± √D)/(2a) with D = b² − 4ac. When D < 0, write √D = i√|D|; the roots are then complex conjugates p ± iq. By the Fundamental Theorem of Algebra there are always exactly two roots, possibly equal. For real coefficients, non-real roots always come in conjugate pairs.

Worked example

Solve 2x² − 3x + 5 = 0.

Try it first, then show the solutionSolution

1. D = 9 − 40 = −31
2. x = (3 ± √(−31))/4 = (3 ± i√31)/4

Answer: x = 3/4 ± (√31/4)i

Equations reducible to quadratic form

Look for an expression that appears together with its square, and substitute y for it: biquadratic equations ax⁴ + bx² + c = 0 (y = x²); exponential equations in aˣ (y = aˣ, remembering y > 0); repeated brackets such as (x² + x)² − 8(x² + x) + 12 = 0 (y = x² + x); reciprocal forms using y = x + 1/x; and radical equations, where you square both sides and then check every root, because squaring can introduce extra roots.

Worked example

Solve 2·4ˣ − 9·2ˣ + 4 = 0.

Try it first, then show the solutionSolution

1. Let y = 2ˣ > 0, so 4ˣ = y²: 2y² − 9y + 4 = 0
2. (2y − 1)(y − 4) = 0, so y = 1/2 or y = 4
3. 2ˣ = 1/2 gives x = −1; 2ˣ = 4 gives x = 2

Answer: x = −1 or x = 2

Nature of roots from the discriminant

For real a, b, c: D > 0: two distinct real roots; if, further, a, b, c are rational and D is a perfect square, the roots are rational; if D is not a perfect square they are irrational and occur as a conjugate pair p ± √q. D = 0: two equal real roots, each −b/(2a). D < 0: two non-real complex conjugate roots. Questions often ask for the value of a parameter making the roots equal, real or complex: write D in terms of the parameter and solve the resulting equation or inequality.

Worked example

Find k so that x² + (k − 3)x + k = 0 has equal roots.

Try it first, then show the solutionSolution

1. D = (k − 3)² − 4k = k² − 10k + 9
2. D = 0: (k − 1)(k − 9) = 0

Answer: k = 1 or k = 9

Sum and product of roots; special root conditions

If α, β are the roots of ax² + bx + c = 0, then ax² + bx + c = a(x − α)(x − β), and comparing coefficients gives α + β = −b/a and αβ = c/a. The equation with roots α, β is x² − (α + β)x + αβ = 0. Special conditions: roots reciprocal (αβ = 1) ⇔ c = a; roots equal in size and opposite in sign ⇔ b = 0; one root the square of the other (roots α, α²) ⇔ b³ + a²c + ac² = 3abc, obtained by cubing α + α² = −b/a and using α³ = c/a.

Worked example

Derive the condition for one root of ax² + bx + c = 0 to be the square of the other, and check it for x² − 6x + 8 = 0.

Try it first, then show the solutionSolution

1. Roots α, α²: α + α² = −b/a and α³ = c/a
2. Cube the first: α³ + α⁶ + 3α³(α + α²) = −b³/a³, i.e. c/a + c²/a² − 3bc/a² = −b³/a³
3. Multiply by a³: a²c + ac² − 3abc = −b³, so b³ + a²c + ac² = 3abc
4. Check (roots 2 and 4 = 2²): b³ + a²c + ac² = −216 + 8 + 64 = −144 and 3abc = 3(1)(−6)(8) = −144 ✓

Answer: b³ + a²c + ac² = 3abc

Symmetric expressions and transformed equations

An expression in α, β that is unchanged when α and β swap is symmetric and can be written using s = α + β and p = αβ only: α² + β² = s² − 2p; (α − β)² = s² − 4p; α³ + β³ = s³ − 3ps; 1/α + 1/β = s/p; α/β + β/α = (s² − 2p)/p. To find the equation whose roots are f(α), f(β), compute the new sum and new product and write x² − (new sum)x + (new product) = 0. This is how you get equations whose roots are α³, β³ or 1/α, 1/β without solving the original.

Worked example

If α, β are the roots of x² − 3x + 1 = 0, find the equation whose roots are α³ and β³.

Try it first, then show the solutionSolution

1. s = 3, p = 1
2. α³ + β³ = s³ − 3ps = 27 − 9 = 18; α³β³ = p³ = 1

Answer: x² − 18x + 1 = 0

Common roots

If a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 share a root α, then α satisfies both. The quickest method is to eliminate x² (multiply to match the x² coefficients and subtract), solve the resulting linear equation for α, and substitute back into either equation. The general condition for exactly one common root is (c₁a₂ − c₂a₁)² = (a₁b₂ − a₂b₁)(b₁c₂ − b₂c₁). Both roots are common when a₁/a₂ = b₁/b₂ = c₁/c₂.

Worked example

Find k if x² − 5x + 6 = 0 and x² + kx − 6 = 0 have a common root.

Try it first, then show the solutionSolution

1. Subtracting: (k + 5)x − 12 = 0, so the common root is 12/(k + 5)
2. The roots of x² − 5x + 6 = 0 are 2 and 3
3. Common root 2: 4 + 2k − 6 = 0, k = 1. Common root 3: 9 + 3k − 6 = 0, k = −1

Answer: k = 1 or k = −1

Sign of a quadratic and its graph

The graph of y = ax² + bx + c is a parabola with axis x = −b/(2a) and vertex (−b/(2a), −D/(4a)). a > 0: opens upwards. If D < 0 it never meets the x-axis, so the quadratic is positive for all real x; if D = 0 it touches the axis and is ≥ 0 everywhere; if D > 0 it is negative only between the roots. a < 0: everything reverses (opens downwards; negative for all x when D < 0; positive only between the roots when D > 0). In short: outside the roots the quadratic has the same sign as a; between the roots it has the opposite sign.

Worked example

State the sign of −x² + 4x − 7 for real x, and of x² − 5x + 4.

Try it first, then show the solutionSolution

1. −x² + 4x − 7: a = −1 < 0, D = 16 − 28 = −12 < 0, so it is negative for every real x
2. x² − 5x + 4 = (x − 1)(x − 4): a > 0, negative for 1 < x < 4, zero at 1 and 4, positive otherwise

Answer: Always negative; negative only between 1 and 4

Maximum and minimum value

Completing the square, ax² + bx + c = a(x + b/(2a))² − D/(4a). If a > 0 the squared term is never negative, so the minimum value is −D/(4a) at x = −b/(2a) and there is no maximum. If a < 0 the maximum is −D/(4a) at x = −b/(2a), with no minimum. The range of the function is [−D/(4a), ∞) or (−∞, −D/(4a)] respectively.

Worked example

Find the maximum value of f(x) = −2x² + 8x − 3 and where it occurs.

Try it first, then show the solutionSolution

1. a = −2 < 0, so a maximum exists at x = −8/(2 × −2) = 2
2. f(2) = −8 + 16 − 3 = 5; check: −D/(4a) = −(64 − 24)/(−8) = 5 ✓

Answer: Maximum 5 at x = 2

Maths in real life: Quadratic Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Equations Reducible to Quadratic, Common Roots and Extreme Values
• Nature of Roots and Relations between the Roots

Quadratic Equations formula sheet

Quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

If D < 0, √D = i√|D|.

Discriminant

$D=b^2-4ac$

D > 0 distinct real; D = 0 equal; D < 0 complex conjugates.

Sum and product

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

Watch the minus sign on the sum.

Equation from roots

$x^2-(\alpha +\beta )x+\alpha \beta =0$

Clear fractions at the end.

Symmetric expressions

${\alpha }^2+{\beta }^2=s^2-2p;{\alpha }^3+{\beta }^3=s^3-3ps;(\alpha -\beta {)}^2=s^2-4p$

s = α + β, p = αβ.

One root the square of the other

$b^3+a^{2}c+a{c}^2=3abc$

Reciprocal roots: c = a.

Common root condition

$(c_1{a}_2-c_2{a}_1{)}^2=(a_1{b}_2-a_2{b}_1)(b_1{c}_2-b_2{c}_1)$

In practice, eliminate x² and solve.

Vertex and extreme value

$x=-\frac{b}{2a},y_{\text{extreme}}=-\frac{D}{4a}$

Minimum if a > 0, maximum if a < 0.

Quadratic Equations: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Quadratic Equations

The question patterns to expect, and the method that works for each.

Tricks to remember Quadratic Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Sum is minus b over a

  Sum takes the sign change (−b/a); product keeps it (c/a).

  Use it when: Sum and product of roots.

• D decides

  Positive: two real; zero: one repeated; negative: complex pair.

  Use it when: Nature of roots.

• Outside same, inside opposite

  Outside the roots a quadratic has the sign of a; between the roots, the opposite sign.

  Use it when: Sign of a quadratic.

• Cube of the sum

  α³ + β³ = s³ − 3ps: 'sum cubed minus three p s'.

  Use it when: Transformed equations.

• Square and check

  Whenever you square both sides of a radical equation, substitute every answer back.

  Use it when: Equations reducible to quadratics.

Quadratic Equations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Quadratic Equations

The slips examiners see most often, and what to write instead.

• ✗Mistake: Dropping the minus sign in the sum of roots

  Students write α + β = b/a.

  Wrong working

  For x² − 5x + 3 = 0, α + β = −5.

  Right working

  α + β = −(−5)/1 = 5.

  ✓Do this: Write −b/a with the actual b in brackets.

  Usually costs: Every later step is wrong.

• ✗Mistake: Keeping a false root of a radical equation

  Squaring both sides adds the roots of the 'negative' version of the equation.

  Wrong working

  √(2x + 7) = x + 2 gives x = 1 or x = −3.

  Right working

  x = −3 gives √1 = −1, false; only x = 1 is a root.

  ✓Do this: Substitute every answer into the original equation.

  Usually costs: 1 mark.

• ✗Mistake: Dividing by x and losing a root

  Students cancel x from both sides of x² = 3x.

  Wrong working

  x² = 3x ⇒ x = 3.

  Right working

  x² − 3x = 0 ⇒ x(x − 3) = 0 ⇒ x = 0 or 3.

  ✓Do this: Factorise instead of cancelling.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting that 2ˣ and similar substitutions must be positive

  Students accept a negative value of y = aˣ.

  Wrong working

  2ˣ = −3 gives x = log₂(−3).

  Right working

  2ˣ > 0 for all real x, so y = −3 is rejected.

  ✓Do this: State the restriction on y when you substitute.

  Usually costs: 1 mark.

• ✗Mistake: Mixing up maximum and minimum

  Students find the vertex but do not check the sign of a.

  Wrong working

  f(x) = 3x² − 12x + 7 has maximum −5.

  Right working

  a = 3 > 0, so −5 is the minimum; there is no maximum.

  ✓Do this: a > 0: cup, minimum; a < 0: cap, maximum.

  Usually costs: 1 mark.

Practice questions on Quadratic Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The quadratic equation whose roots are 2 and −5 is:
   (a) x² − 3x − 10 = 0
   (b) x² + 3x − 10 = 0
   (c) x² + 3x + 10 = 0
   (d) x² − 7x + 10 = 0

   Hint

   x² − (sum)x + product.

   Answer

   (b): sum −3, product −10

2. Q2 · 1 mark · MCQExam level

   The roots of x² − 4x + 4 = 0 are:
   (a) real and distinct
   (b) real and equal
   (c) non-real
   (d) irrational

   Hint

   Find D.

   Answer

   (b): D = 0

3. Q3 · 1 mark · Very short answerExam level

   Write the sum and product of the roots of 5x² + 3x − 2 = 0.

   Hint

   −b/a and c/a.

   Answer

   Sum −3/5; product −2/5

4. Q4 · 1 mark · Very short answerExam level

   Solve x² + 9 = 0.

   Hint

   x² = −9.

   Answer

   x = ±3i

5. Q5 · 2 marks · Short answerExam level

   Solve x² − 2x + 5 = 0.

   Hint

   D = −16.

   Answer

   x = (2 ± 4i)/2 = 1 ± 2i

6. Q6 · 2 marks · Short answerHarder exam level

   Solve √(2x + 7) = x + 2.

   Hint

   Square, then check.

   Answer

   2x + 7 = x² + 4x + 4 ⇒ x² + 2x − 3 = 0 ⇒ x = 1 or −3. x = −3 fails (√1 ≠ −1), so x = 1.

7. Q7 · 2 marks · Short answerHarder exam level

   If one root of x² − kx + 12 = 0 is three times the other, find k.

   Hint

   Roots α and 3α.

   Answer

   3α² = 12 ⇒ α = ±2; k = α + 3α = 4α = ±8

8. Q8 · 3 marks · Short answerHarder exam level

   Solve (x² + x)² − 8(x² + x) + 12 = 0.

   Hint

   Let y = x² + x.

   Answer

   y² − 8y + 12 = 0 ⇒ y = 2 or 6. x² + x − 2 = 0 ⇒ x = 1, −2; x² + x − 6 = 0 ⇒ x = 2, −3.

9. Q9 · 3 marks · Short answerHarder exam level

   If α, β are the roots of x² + 2x + 4 = 0, form the equation whose roots are α² and β².

   Hint

   s = −2, p = 4.

   Answer

   α² + β² = 4 − 8 = −4; α²β² = 16; equation x² + 4x + 16 = 0

10. Q10 · 2 marks · Short answerHarder exam level

    Find the minimum value of f(x) = 3x² − 12x + 7 and the value of x at which it occurs.

    Hint

    x = −b/(2a).

    Answer

    x = 2; f(2) = 12 − 24 + 7 = −5

11. Q11 · 3 marks · Short answerHarder exam level

    Prove that the roots of x² − 2(a + b)x + 2(a² + b²) = 0 are non-real unless a = b, where a, b are real.

    Hint

    Find D/4.

    Answer

    D/4 = (a + b)² − 2(a² + b²) = −(a² − 2ab + b²) = −(a − b)². This is negative unless a = b, when D = 0 and the roots are real and equal.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Find p if x² + px + 9 = 0 has equal roots. [2]
    (b) For p = 10, solve the equation and state the nature of the roots. [2]
    (c) For p = 4, state the sign of x² + 4x + 9 for all real x, with a reason. [1]

    Hint

    D = p² − 36.

    Answer

    (a) p² − 36 = 0 ⇒ p = ±6. (b) x² + 10x + 9 = (x + 1)(x + 9) ⇒ x = −1, −9: real, rational, distinct (D = 64, a perfect square). (c) D = 16 − 36 < 0 and a > 0, so it is positive for all real x.

13. Q13 · 3 marks · Short answerHOTS / competency

    If x² + px + q = 0 and x² + qx + p = 0 (p ≠ q) have a common root, prove that p + q + 1 = 0.

    Hint

    Subtract the equations.

    Answer

    (p − q)x + (q − p) = 0 ⇒ x = 1 (as p ≠ q). Substituting: 1 + p + q = 0.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the set of values taken by y = (x² − x + 1)/(x² + x + 1) for real x.

    Hint

    Make it a quadratic in x and demand real roots.

    Answer

    (y − 1)x² + (y + 1)x + (y − 1) = 0 must have real x: (y + 1)² − 4(y − 1)² ≥ 0 ⇒ (3 − y)(3y − 1) ≥ 0 ⇒ 1/3 ≤ y ≤ 3 (y = 1 occurs at x = 0).

15. Q15 · 5 marks · Long answerHarder exam level

    (a) Show that 2x² − 4x + 5 > 0 for all real x. [1]
    (b) If α, β are the roots of 2x² − 4x + 5 = 0, find α² + β². [2]
    (c) Form the equation whose roots are α + 1/β and β + 1/α. [2]

    Hint

    s = 2, p = 5/2.

    Answer

    (a) a > 0 and D = 16 − 40 < 0. (b) 4 − 5 = −1. (c) Sum = s + s/p = 2 + 4/5 = 14/5; product = p + 2 + 1/p = 5/2 + 2 + 2/5 = 49/10. Equation: 10x² − 28x + 49 = 0.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Quadratic formula?Show answer

   x = (−b ± √(b² − 4ac))/(2a).

2. 2.D < 0 means?Show answer

   Two non-real complex conjugate roots.

3. 3.When are the roots rational?Show answer

   a, b, c rational and D a perfect square.

4. 4.α³ + β³ in terms of s and p?Show answer

   s³ − 3ps.

5. 5.Condition for reciprocal roots?Show answer

   c = a.

6. 6.Condition for one root the square of the other?Show answer

   b³ + a²c + ac² = 3abc.

7. 7.Extreme value of ax² + bx + c?Show answer

   −D/(4a) at x = −b/(2a): minimum if a > 0, maximum if a < 0.

8. 8.Sign of ax² + bx + c when D < 0?Show answer

   Same as the sign of a, for every real x.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you derive the quadratic formula?
• ☐Can you solve a quadratic whose roots are complex?
• ☐Can you solve a biquadratic, an exponential and a radical equation that reduce to quadratics?
• ☐Can you find k for equal roots?
• ☐Can you write α² + β², α³ + β³ and (α − β)² in terms of s and p?
• ☐Can you form the equation whose roots are α³, β³?
• ☐Can you find the condition for a common root?
• ☐Can you find the maximum or minimum and the sign of any quadratic?

Quadratic Equations: frequently asked questions