Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers modelling, factorisation and discriminant/formula use. It gives selected reasoning and examples rather than every contextual problem or a full derivation of the quadratic formula.
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LESSON 1
Build and recognise a quadratic equation
Convert a product situation into ax² + bx + c = 0.
A square term x² comes from multiplying x by x.
Define the variable
Let a rectangle have width x metres and length x + 2 metres. Its area is 48 m².
Model the area
Area = length × width, so x(x + 2) = 48.
Write standard form
Expand and bring all terms to one side: x² + 2x − 48 = 0.
Keep the real situation
A solution used as a length must be positive. Algebraic roots must be checked against this condition.
Worked example: The rectangle equation factors as (x + 8)(x − 6) = 0. The roots are −8 and 6, but only width 6 m is positive. Length 8 m gives area 48 m².
Watch for this: A quadratic requires a nonzero x² coefficient after simplification. A negative algebraic root is not automatically a valid physical length.
Discuss: Model two consecutive positive integers whose product is 56, then identify the positive pair.
LESSON 2
Solve by factorisation
Use the zero-product rule to retain both roots.
A product can be zero even when only one factor is zero.
Put zero on one side
For x² = 9x − 20, rewrite as x² − 9x + 20 = 0.
Factor the expression
Two numbers with sum 9 and product 20 are 4 and 5. Thus (x − 4)(x − 5) = 0.
Separate two cases
Either x − 4 = 0 or x − 5 = 0. Keep both roots: 4 and 5.
Verify the roots
For x = 4, 16 = 36 − 20. For x = 5, 25 = 45 − 20. Both satisfy the original equation.
Worked example: Solve 2x² − 10x = 0. Factor 2x(x − 5) = 0. Thus x = 0 or x = 5. Checking gives 0 and 50 − 50 respectively.
Watch for this: Do not divide by x when x might be zero; doing so can lose a root. For x(x − 5) = 0, both 0 and 5 are roots.
Discuss: Make a quadratic whose roots are 0 and −4. Explain why dividing by x would lose the zero root.
LESSON 3
Use the discriminant and formula
Decide the real-root type and calculate roots when D is nonnegative.
The square of any real number is nonnegative.
Identify the signed coefficients
For 2x² − 7x + 3 = 0, a = 2, b = −7, c = 3.
Calculate D
D = (−7)² − 4 × 2 × 3 = 49 − 24 = 25. Positive D means two distinct real roots.
Use both signs
x = (7 ± 5)/4. The plus sign gives 3; the minus sign gives 1/2.
Interpret other signs
If D = 0 the roots coincide; if D < 0 there are no real roots. This does not mean the equation has no roots in every number system.
Worked example: For x² + 4x + 4 = 0, D = 16 − 16 = 0. The formula gives x = −4/2 = −2 twice, matching (x + 2)² = 0.
Watch for this: The whole numerator −b ± √D is divided by 2a. Squaring a negative b produces a positive number.
Discuss: Compare x² − 4x + c = 0 for c = 3, 4 and 5. Use D to explain the three different real-root cases.
Original learning companion, not full textbook exercise solutions. Covers modelling, factorisation and discriminant/formula use. It gives selected reasoning and examples rather than every contextual problem or a full derivation of the quadratic formula.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
