This MCQ module is based on: Quadratic Formula and Exercises
TOPIC 10 OF 35
Quadratic Formula and Exercises
🎓 Class 10
Mathematics
CBSE
Theory
Ch 4 — Quadratic Equations
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Quadratic Formula and Exercises
Targeting Class 10 level in Algebra, with Intermediate difficulty.
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4.4 The Quadratic Formula (Solution by Completing the Square)
Factorisation is quick when the middle term splits into rational pieces — but many quadratic equations resist this. For every quadratic equation \(ax^2 + bx + c = 0\) with \(a \ne 0\), we can derive a universal formula by completing the square.
Derivation of the Quadratic Formula
Start with \(ax^2 + bx + c = 0\). Divide by \(a\):
\[ x^2 + \tfrac{b}{a}x + \tfrac{c}{a} = 0 \]
Rewrite by adding and subtracting \(\left(\tfrac{b}{2a}\right)^2\):
\[ \left(x + \tfrac{b}{2a}\right)^2 - \tfrac{b^2}{4a^2} + \tfrac{c}{a} = 0 \]
\[ \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2} \]
If \(b^2 - 4ac \ge 0\), take square roots:
\[ x + \tfrac{b}{2a} = \pm \tfrac{\sqrt{b^2 - 4ac}}{2a} \quad\Longrightarrow\quad x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This is the famous quadratic formula, also known in India as the Sridharacharya formula.
4.4.1 Discriminant and Nature of Roots
The quantity \(D = b^2 - 4ac\) that appears under the root sign is called the discriminant?. Its sign decides everything about the roots.
Nature of Roots by Discriminant
(i) \(D > 0\): two distinct real roots — \(x = \dfrac{-b + \sqrt{D}}{2a}\) and \(x = \dfrac{-b - \sqrt{D}}{2a}\).(ii) \(D = 0\): two equal real roots — \(x = -\dfrac{b}{2a}\) (a repeated root).
(iii) \(D < 0\): no real roots; the equation has complex-conjugate roots (studied later).
Worked Example 7 — Applying the Formula
Find the discriminant and hence the nature of roots of \(2x^2 - 4x + 3 = 0\).
\(a = 2, b = -4, c = 3\). Discriminant \(D = (-4)^2 - 4(2)(3) = 16 - 24 = -8\).
Since \(D < 0\), the equation has no real roots.
Worked Example 8 — Pole Between Gates
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, find the distances from the two gates.
Fig 4.2 — Pole P on the boundary; \(\angle APB = 90°\) (angle in a semicircle).
Let pole-to-gate-B distance \(BP = x\). Then \(AP = x + 7\) (since \(AP - BP = 7\)).
Since AB is a diameter (13 m), \(\angle APB = 90°\) (angle in a semicircle). By Pythagoras: \(AP^2 + BP^2 = AB^2\).
\((x + 7)^2 + x^2 = 13^2 \Rightarrow x^2 + 14x + 49 + x^2 = 169 \Rightarrow 2x^2 + 14x - 120 = 0 \Rightarrow x^2 + 7x - 60 = 0\).
\(D = 49 + 240 = 289 > 0\). Formula: \(x = \dfrac{-7 \pm 17}{2}\); take positive root: \(x = 5\).
Yes, it is possible. Pole is 5 m from gate B and 12 m from gate A.
Worked Example 9 — Discriminant for Equality
Find \(k\) so that \(2x^2 + kx + 3 = 0\) has two equal roots.
For equal roots: \(D = 0 \Rightarrow k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm 2\sqrt{6}\).
Worked Example 10 — Side of a Rectangle
Is it possible to design a rectangular mango grove of length twice its breadth, with area 800 m²?
Let breadth = \(x\). Length = \(2x\). Area: \(2x \cdot x = 800 \Rightarrow x^2 = 400 \Rightarrow x = 20\).
Yes — a 20 m × 40 m grove works. The discriminant of \(2x^2 - 800 = 0\) (writing as \(2x^2 + 0 \cdot x - 800 = 0\)) is \(0 + 6400 = 6400 > 0\), confirming real dimensions exist.
Activity: Discriminant Detective
Materials: paper, pen.
- Without solving, predict the nature of roots for: (i) \(2x^2 - 3x + 5 = 0\) (ii) \(3x^2 - 4\sqrt{3}x + 4 = 0\) (iii) \(2x^2 - 6x + 3 = 0\).
- Now compute \(D\) for each and verify.
- Make up two quadratics — one with equal roots, one with distinct real roots — and ask your partner to classify them.
(i) \(D = 9 - 40 = -31 < 0\): no real roots.
(ii) \(D = 48 - 48 = 0\): equal real roots \(x = \tfrac{4\sqrt{3}}{6} = \tfrac{2\sqrt{3}}{3}\).
(iii) \(D = 36 - 24 = 12 > 0\): two distinct real roots \(x = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}\).
(ii) \(D = 48 - 48 = 0\): equal real roots \(x = \tfrac{4\sqrt{3}}{6} = \tfrac{2\sqrt{3}}{3}\).
(iii) \(D = 36 - 24 = 12 > 0\): two distinct real roots \(x = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}\).
Interactive: Quadratic Formula Solver
Enter a, b, c and click Solve.
Exercise 4.3
Q1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
(i) \(2x^2 - 3x + 5 = 0\) (ii) \(3x^2 - 4\sqrt{3}x + 4 = 0\) (iii) \(2x^2 - 6x + 3 = 0\)
(i) \(2x^2 - 3x + 5 = 0\) (ii) \(3x^2 - 4\sqrt{3}x + 4 = 0\) (iii) \(2x^2 - 6x + 3 = 0\)
See Activity reveal above. (i) no real roots; (ii) equal roots \(x = 2\sqrt{3}/3\); (iii) two distinct real roots \(x = (3 \pm \sqrt{3})/2\).
Q2. Find the values of \(k\) for each of the following quadratic equations, so that they have two equal roots:
(i) \(2x^2 + kx + 3 = 0\) (ii) \(kx(x - 2) + 6 = 0\)
(i) \(2x^2 + kx + 3 = 0\) (ii) \(kx(x - 2) + 6 = 0\)
(i) \(D = k^2 - 24 = 0 \Rightarrow k = \pm 2\sqrt{6}\).
(ii) Rewrite: \(kx^2 - 2kx + 6 = 0\). \(D = 4k^2 - 24k = 4k(k - 6) = 0 \Rightarrow k = 0\) or \(k = 6\). Reject \(k = 0\) (not quadratic), so \(\mathbf{k = 6}\).
(ii) Rewrite: \(kx^2 - 2kx + 6 = 0\). \(D = 4k^2 - 24k = 4k(k - 6) = 0 \Rightarrow k = 0\) or \(k = 6\). Reject \(k = 0\) (not quadratic), so \(\mathbf{k = 6}\).
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its dimensions.
From Worked Example 10 — yes. Breadth 20 m, length 40 m.
Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48.
Let ages be \(x\) and \(20 - x\). Four years ago: \((x - 4)(16 - x) = 48 \Rightarrow 16x - x^2 - 64 + 4x = 48 \Rightarrow -x^2 + 20x - 112 = 0 \Rightarrow x^2 - 20x + 112 = 0\). \(D = 400 - 448 = -48 < 0\). Not possible — the situation has no real-valued age pair.
Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Perimeter: \(2(l + b) = 80 \Rightarrow l + b = 40\). Area: \(lb = 400\). So \(l, b\) are roots of \(t^2 - 40t + 400 = 0\). \(D = 1600 - 1600 = 0 \Rightarrow t = 20\). Yes — a 20 m × 20 m square park.
Competency-Based Questions
Scenario: A small factory produces gadgets. The daily profit (in ₹ thousand) as a function of the number of gadgets \(x\) made per day is modelled by \(P(x) = -x^2 + 14x - 40\). The manager wants to know break-even points (profit = 0) and the maximum profit.
Q1. Find the break-even values of \(x\) using the quadratic formula.
L3 Apply\(-x^2 + 14x - 40 = 0 \Rightarrow x^2 - 14x + 40 = 0\). \(D = 196 - 160 = 36\). \(x = (14 \pm 6)/2 = 10 \text{ or } 4\). Break-even at x = 4 and x = 10 gadgets.
Q2. Analyse: interpret the discriminant's sign for this profit equation — what does \(D > 0\) mean in business terms?
L4 Analyse\(D > 0\) means the parabolic profit curve crosses the \(x\)-axis at two distinct values — there exists a profitable range of production \((4, 10)\). If \(D = 0\), only a single touching point (borderline profitability); \(D < 0\), the curve never reaches zero: the business is always in loss or always in profit (sign depends on leading coefficient).
Q3. Evaluate: a consultant suggests producing 7 gadgets/day. Is this optimal? Justify using the vertex of the parabola.
L5 EvaluateVertex \(x = -b/(2a) = -14/(-2) = 7\). Indeed optimal. \(P(7) = -49 + 98 - 40 = 9\) → max profit Rs. 9,000/day. The consultant is correct.
Q4. Design: propose a new profit function \(P(x) = -x^2 + bx + c\) for which (a) break-even occurs at \(x = 3\) and \(x = 12\), and (b) maximum profit equals Rs. 20,000. Find \(b\) and \(c\).
L6 CreateSum of roots = \(b = 3 + 12 = 15\). Product = \(-c/(-1) = c = 3 \times 12 = 36\). Check vertex: \(x^* = 7.5\); \(P(7.5) = -56.25 + 112.5 - 36 = 20.25\) → Rs. 20,250. Close to Rs. 20,000 but not exact. To make exact: scale leading coefficient. \(b = 15,\; c = 36\) gives the required break-even points with max profit ≈ Rs. 20,250.
Assertion–Reason Questions
Assertion (A): The equation \(x^2 + x + 1 = 0\) has no real roots.
Reason (R): Discriminant \(D = 1 - 4 = -3 < 0\).
Reason (R): Discriminant \(D = 1 - 4 = -3 < 0\).
(a).
Assertion (A): If \(b^2 = 4ac\), the quadratic \(ax^2 + bx + c = 0\) has two equal real roots.
Reason (R): The quadratic formula with \(D = 0\) gives \(x = -b/(2a)\), a single value.
Reason (R): The quadratic formula with \(D = 0\) gives \(x = -b/(2a)\), a single value.
(a).
Assertion (A): The equation \(x^2 - 4x + 5 = 0\) can be solved by factorisation over the real numbers.
Reason (R): Every quadratic can be written as a product of two real linear factors.
Reason (R): Every quadratic can be written as a product of two real linear factors.
Both false. \(D = 16 - 20 = -4 < 0\), so no real factorisation. R is false — only when \(D \ge 0\) does real factorisation exist.
4.5 Summary
- A quadratic equation in \(x\) has the standard form \(a x^2 + b x + c = 0\), \(a \ne 0\).
- A real number \(\alpha\) is a root of the quadratic if \(a\alpha^2 + b\alpha + c = 0\); equivalently, \(\alpha\) is a zero of the polynomial \(ax^2 + bx + c\). A quadratic has at most two roots.
- If the LHS factorises as a product of two linear factors, the roots are found by setting each factor equal to zero — the factorisation method.
- Completing the square on the general quadratic yields the quadratic formula (Sridharacharya): \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), valid when \(b^2 - 4ac \ge 0\).
- The discriminant \(D = b^2 - 4ac\) determines the nature of roots: \(D > 0\) two distinct real roots; \(D = 0\) two equal real roots; \(D < 0\) no real roots.
Key Terms
Standard form? · Factorisation? · Completing the square? · Discriminant? · Quadratic formula?
Frequently Asked Questions
What is the quadratic formula?
For ax^2 + bx + c = 0 (a != 0), the roots are x = (-b ± sqrt(b^2 - 4ac)) / (2a). This formula always works when roots are real. NCERT Class 10 Maths Chapter 4 derives this from completing the square.
What is the discriminant?
The discriminant D = b^2 - 4ac determines the nature of roots. If D > 0, two distinct real roots. If D = 0, one repeated real root. If D < 0, no real roots (two complex). NCERT Class 10 Chapter 4 uses D extensively.
How do you determine the nature of roots without solving?
Compute D = b^2 - 4ac. The sign of D tells you: positive means distinct real roots, zero means equal real roots, negative means no real roots. NCERT Class 10 Maths Chapter 4 emphasises this shortcut.
Solve 2x^2 - 4x - 3 = 0 using the quadratic formula.
a=2, b=-4, c=-3. D = 16 - 4(2)(-3) = 16 + 24 = 40. x = (4 ± sqrt(40))/4 = (4 ± 2sqrt(10))/4 = (2 ± sqrt(10))/2. NCERT Class 10 Chapter 4 shows such applications.
What does it mean if discriminant is zero?
D = 0 means the quadratic has exactly one real root (a repeated root), x = -b/(2a). The parabola just touches the x-axis. Example: x^2 - 6x + 9 = (x-3)^2 = 0 has double root x = 3. NCERT Class 10 Maths Chapter 4 explains.
Why learn quadratic formula if factorisation works?
The quadratic formula works for every quadratic, including those with irrational or no real roots. Factorisation is faster when applicable but fails for many cases. Together they form a complete toolkit. NCERT Class 10 Chapter 4 teaches both.
Frequently Asked Questions — Quadratic Equations
What is Quadratic Formula and Exercises in NCERT Class 10 Mathematics?
Quadratic Formula and Exercises is a key concept covered in NCERT Class 10 Mathematics, Chapter 4: Quadratic Equations. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Quadratic Formula and Exercises step by step?
To solve problems on Quadratic Formula and Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 4: Quadratic Equations?
The essential formulas of Chapter 4 (Quadratic Equations) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Quadratic Formula and Exercises important for the Class 10 board exam?
Quadratic Formula and Exercises is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Quadratic Formula and Exercises?
Common mistakes in Quadratic Formula and Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Quadratic Formula and Exercises?
End-of-chapter NCERT exercises for Quadratic Formula and Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 4, and solve at least one previous-year board paper to consolidate your understanding.
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