NCERT Exemplar Class 10 Maths Chapter 4 Quadratic Equations Solutions (CBSE 2026-27)
This post covers in-syllabus NCERT Exemplar questions strictly based on the rationalised curriculum: identifying quadratic equations, finding roots by factorisation and quadratic formula, discriminant and nature of roots, and situational real-life word problems. Questions based on the omitted Method of Completing the Square have been pruned.
Section A: Multiple Choice Questions (MCQs)
Simplifying option (C):
\[(x + 2)(x – 1) = x^2 + x – 2\] \[x^2 + x – 2 = x^2 – 2x – 3 \implies 3x + 1 = 0\]Since the degree of this simplified equation is \(1\), it is a linear equation and not a quadratic equation.
Expanding option (D):
\[(x – 1)^3 = x^3 – 3x^2 + 3x – 1\] \[x^3 – x^2 = x^3 – 3x^2 + 3x – 1\] \[-x^2 + 3x^2 – 3x + 1 = 0 \implies 2x^2 – 3x + 1 = 0\]Since the degree of the equation is \(2\) and \(a = 2 \neq 0\), it is a quadratic equation.
Expanding the equation in option (C):
\[(2x^2 + 2\sqrt{6}x + 3) + x^2 = 3x^2 – 5x\] \[3x^2 + 2\sqrt{6}x + 3 = 3x^2 – 5x \implies (2\sqrt{6} + 5)x + 3 = 0\]The \(x^2\) terms cancel out, leaving a linear equation. Thus, (C) is not a quadratic equation.
Substituting \(x = 2\) into the equation in option (C):
\[2(2)^2 – 7(2) + 6 = 2(4) – 14 + 6 = 8 – 14 + 6 = 0\]Since \(x = 2\) satisfies the equation, \(2\) is a root of \(2x^2 – 7x + 6 = 0\).
Since \(\frac{1}{2}\) is a root, substituting \(x = \frac{1}{2}\) into the equation:
\[\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) – \frac{5}{4} = 0\] \[\frac{1}{4} + \frac{k}{2} – \frac{5}{4} = 0 \implies \frac{k}{2} – \frac{4}{4} = 0 \implies \frac{k}{2} = 1 \implies k = 2\]The sum of roots of a quadratic equation \(ax^2 + bx + c = 0\) is \(-\frac{b}{a}\).
For option (B), \(-x^2 + 3x – 3 = 0\): \(a = -1\), \(b = 3\).
\[\text{Sum of roots} = -\frac{b}{a} = -\frac{3}{-1} = 3\]For equal roots, the discriminant must be zero: \(D = b^2 – 4ac = 0\).
\[(-k)^2 – 4(2)(k) = 0 \implies k^2 – 8k = 0 \implies k(k – 8) = 0\]Therefore, \(k = 0\) or \(k = 8\).
Calculating the discriminant with \(a = 2, b = -\sqrt{5}, c = 1\):
\[D = b^2 – 4ac = (-\sqrt{5})^2 – 4(2)(1) = 5 – 8 = -3\]Since \(D < 0\), the equation has no real roots.
For option (B), \(x^2 + x – 5 = 0\): \(a = 1, b = 1, c = -5\).
\[D = b^2 – 4ac = (1)^2 – 4(1)(-5) = 1 + 20 = 21 > 0\]Since \(D > 0\), the equation has two distinct real roots.
For option (A), \(x^2 – 4x + 3\sqrt{2} = 0\): \(a = 1, b = -4, c = 3\sqrt{2}\).
\[D = (-4)^2 – 4(1)(3\sqrt{2}) = 16 – 12\sqrt{2}\]Since \(\sqrt{2} \approx 1.414\), \(12\sqrt{2} \approx 16.97 > 16\). Thus, \(D = 16 – 16.97 < 0\). The equation has no real roots.
Expanding the equation:
\[(x^4 + 2x^2 + 1) – x^2 = 0 \implies x^4 + x^2 + 1 = 0\]Let \(y = x^2\). The equation becomes \(y^2 + y + 1 = 0\). The discriminant is:
\[D = 1^2 – 4(1)(1) = -3 < 0\]Since \(D < 0\), there are no real values for \(y\), which means there are no real roots for \(x\).
Section B: Short Answer Questions with Reasoning
Expanding and simplifying the given equation:
\[(x^2 – 2x + 1) + 2x + 2 = 0 \implies x^2 + 3 = 0\]Here, \(a = 1, b = 0, c = 3\). The discriminant is:
\[D = b^2 – 4ac = 0^2 – 4(1)(3) = -12 < 0\]Since \(D < 0\), the equation has no real roots.
By the Fundamental Theorem of Algebra, a polynomial equation of degree \(n\) has exactly \(n\) complex roots (counting multiplicity). A quadratic equation has degree \(2\), so it can have at most \(2\) distinct real roots.
Let the quadratic equation be \(ax^2 + bx + c = 0\). If \(a\) and \(c\) have opposite signs, their product \(ac < 0\).
The discriminant is \(D = b^2 – 4ac\). Since \(b^2 \ge 0\) and \(-4ac > 0\),
\[D = b^2 – 4ac > 0\]Because \(D > 0\), the quadratic equation will always have two distinct real roots.
Here, \(b = 0\), and \(a\) and \(c\) have the same sign (\(ac > 0\)).
The discriminant is:
\[D = b^2 – 4ac = 0 – 4ac = -4ac\]Since \(ac > 0\), \(-4ac < 0\), meaning \(D < 0\). Therefore, the equation has no real roots.
Substitute \(x = 0.2\) into the equation:
\[(0.2)^2 – 0.4 = 0.04 – 0.4 = -0.36 \neq 0\]Since \(x = 0.2\) does not satisfy the equation, it is not a root.
Substituting \(b = 0\) into \(x^2 + bx + c = 0\):
\[x^2 + c = 0 \implies x^2 = -c\]Since \(c < 0\), \(-c > 0\). Taking the square root on both sides:
\[x = \pm\sqrt{-c}\]The two roots are \(+\sqrt{-c}\) and \(-\sqrt{-c}\), which are numerically equal and opposite in sign.
Section C: Finding Roots by Factorisation & Quadratic Formula
Here, \(a = 2, b = -\sqrt{5}, c = -2\). Computing the discriminant:
\[D = b^2 – 4ac = (-\sqrt{5})^2 – 4(2)(-2) = 5 + 16 = 21\]Using the quadratic formula \(x = \frac{-b \pm \sqrt{D}}{2a}\):
\[x = \frac{-(-\sqrt{5}) \pm \sqrt{21}}{2(2)} = \frac{\sqrt{5} \pm \sqrt{21}}{4}\]Roots: \(\frac{\sqrt{5} + \sqrt{21}}{4}\) and \(\frac{\sqrt{5} – \sqrt{21}}{4}\).
We split the middle term: product \(= 6 \times (-2) = -12\), sum \(= -\sqrt{2}\). Choose \(-3\sqrt{2}\) and \(2\sqrt{2}\):
\[6x^2 – 3\sqrt{2}x + 2\sqrt{2}x – 2 = 0\] \[3x(2x – \sqrt{2}) + \sqrt{2}(2x – \sqrt{2}) = 0\] \[(3x + \sqrt{2})(2x – \sqrt{2}) = 0\]Setting each factor to zero:
\[3x + \sqrt{2} = 0 \implies x = -\frac{\sqrt{2}}{3}\] \[2x – \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\]Roots: \(-\frac{\sqrt{2}}{3}\) and \(\frac{\sqrt{2}}{2}\).
Here, \(a = 2, b = -3, c = -5\):
\[D = b^2 – 4ac = (-3)^2 – 4(2)(-5) = 9 + 40 = 49\]Applying the quadratic formula:
\[x = \frac{-(-3) \pm \sqrt{49}}{2(2)} = \frac{3 \pm 7}{4}\] \[x_1 = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}, \quad x_2 = \frac{3 – 7}{4} = \frac{-4}{4} = -1\]Roots: \(\frac{5}{2}\) and \(-1\).
Here, \(a = 1, b = 2\sqrt{2}, c = -6\):
\[D = (2\sqrt{2})^2 – 4(1)(-6) = 8 + 24 = 32\]Applying the quadratic formula:
\[x = \frac{-2\sqrt{2} \pm \sqrt{32}}{2(1)} = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{2}\] \[x_1 = \frac{-2\sqrt{2} + 4\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}\] \[x_2 = \frac{-2\sqrt{2} – 4\sqrt{2}}{2} = \frac{-6\sqrt{2}}{2} = -3\sqrt{2}\]Roots: \(\sqrt{2}\) and \(-3\sqrt{2}\).
Product \(= 3\sqrt{2} \times (-\sqrt{2}) = -6\), sum \(= -5\). We choose \(-6\) and \(1\):
\[3\sqrt{2}x^2 – 6x + x – \sqrt{2} = 0\] \[3\sqrt{2}x(x – \sqrt{2}) + 1(x – \sqrt{2}) = 0\] \[(3\sqrt{2}x + 1)(x – \sqrt{2}) = 0\]Equating each factor to zero:
\[3\sqrt{2}x + 1 = 0 \implies x = -\frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6}\] \[x – \sqrt{2} = 0 \implies x = \sqrt{2}\]Roots: \(-\frac{\sqrt{2}}{6}\) and \(\sqrt{2}\).
Product \(= 3 \times (-10) = -30\), sum \(= 5\sqrt{5}\). Choose \(6\sqrt{5}\) and \(-\sqrt{5}\) (since \(6\sqrt{5} \times (-\sqrt{5}) = -30\)):
\[3x^2 + 6\sqrt{5}x – \sqrt{5}x – 10 = 0\] \[3x(x + 2\sqrt{5}) – \sqrt{5}(x + 2\sqrt{5}) = 0\] \[(3x – \sqrt{5})(x + 2\sqrt{5}) = 0\]Setting each factor to zero:
\[3x – \sqrt{5} = 0 \implies x = \frac{\sqrt{5}}{3}\] \[x + 2\sqrt{5} = 0 \implies x = -2\sqrt{5}\]Roots: \(\frac{\sqrt{5}}{3}\) and \(-2\sqrt{5}\).
Section D: Real-Life Contextual Word Problems
Let her actual marks be \(x\). According to the problem condition:
\[9(x + 10) = x^2 \implies 9x + 90 = x^2\] \[x^2 – 9x – 90 = 0\]Factoring by splitting the middle term (\(-15 \times 6 = -90\)):
\[x^2 – 15x + 6x – 90 = 0 \implies (x – 15)(x + 6) = 0\]So, \(x = 15\) or \(x = -6\).
Since test marks cannot be negative, \(x \neq -6\). Hence, Ajita scored \(15\) marks.
Let the original average speed be \(x\text{ km/h}\). Time taken for both portions:
\[\frac{63}{x} + \frac{72}{x + 6} = 3\]Dividing the entire equation by \(3\):
\[\frac{21}{x} + \frac{24}{x + 6} = 1\] \[\frac{21(x + 6) + 24x}{x(x + 6)} = 1 \implies 21x + 126 + 24x = x^2 + 6x\] \[45x + 126 = x^2 + 6x \implies x^2 – 39x – 126 = 0\]Factoring (\(-42 \times 3 = -126\)):
\[(x – 42)(x + 3) = 0 \implies x = 42 \quad \text{or} \quad x = -3\]Since speed cannot be negative, \(x = 42\). The original average speed is \(42\text{ km/h}\).
Let the natural number be \(n\). According to the statement:
\[n^2 – 84 = 3(n + 8)\] \[n^2 – 84 = 3n + 24 \implies n^2 – 3n – 108 = 0\]Factoring the quadratic equation (\(-12 \times 9 = -108\)):
\[(n – 12)(n + 9) = 0 \implies n = 12 \quad \text{or} \quad n = -9\]Since \(n\) must be a natural number (\(n \in \mathbb{N}\)), \(n \neq -9\). The required number is \(12\).
Let the natural number be \(x\):
\[x + 12 = \frac{160}{x}\] \[x(x + 12) = 160 \implies x^2 + 12x – 160 = 0\]Factoring the equation (\(20 \times (-8) = -160\)):
\[(x + 20)(x – 8) = 0 \implies x = 8 \quad \text{or} \quad x = -20\]Since \(x\) is a natural number, \(x \neq -20\). Hence, the required number is \(8\).
Let the original speed be \(x\text{ km/h}\). Time difference \(= 48\text{ minutes} = \frac{48}{60} = \frac{4}{5}\text{ hours}\):
\[\frac{360}{x} – \frac{360}{x + 5} = \frac{4}{5}\]Dividing by \(4\):
\[\frac{90}{x} – \frac{90}{x + 5} = \frac{1}{5} \implies 90\left(\frac{x + 5 – x}{x(x + 5)}\right) = \frac{1}{5}\] \[\frac{450}{x^2 + 5x} = \frac{1}{5} \implies x^2 + 5x = 2250 \implies x^2 + 5x – 2250 = 0\]Factoring (\(50 \times (-45) = -2250\)):
\[(x + 50)(x – 45) = 0 \implies x = 45 \quad \text{or} \quad x = -50\]Speed cannot be negative, so \(x = 45\). The original speed of the train is \(45\text{ km/h}\).
Let Zeba’s present age be \(x\) years. According to the condition:
\[(x – 5)^2 = 5x + 11\] \[x^2 – 10x + 25 = 5x + 11 \implies x^2 – 15x + 14 = 0\]Factoring (\(-14 \times (-1) = 14\)):
\[(x – 14)(x – 1) = 0 \implies x = 14 \quad \text{or} \quad x = 1\]If \(x = 1\), her age \(5\) years ago \((1 – 5 = -4)\) is impossible. Therefore, \(x = 14\). Zeba’s present age is \(14\text{ years}\).
Let the grass border surrounding the pond have a uniform width of \(x\) metres.
Total area of lawn \(= 50 \times 40 = 2000\text{ m}^2\).
Area of pond \(= \text{Total area} – \text{Grass area} = 2000 – 1184 = 816\text{ m}^2\).
Dimensions of the pond are \((50 – 2x)\text{ m}\) and \((40 – 2x)\text{ m}\):
\[(50 – 2x)(40 – 2x) = 816\] \[2000 – 100x – 80x + 4x^2 = 816 \implies 4x^2 – 180x + 1184 = 0\]Dividing by \(4\):
\[x^2 – 45x + 296 = 0\]Factoring (\(-37 \times (-8) = 296\)):
\[(x – 8)(x – 37) = 0 \implies x = 8 \quad \text{or} \quad x = 37\]If \(x = 37\), the breadth becomes \(40 – 2(37) < 0\), which is impossible. Thus, \(x = 8\text{ m}\).
Dimensions of pond:
\[\text{Length} = 50 – 2(8) = 34\text{ m}\] \[\text{Breadth} = 40 – 2(8) = 24\text{ m}\]From \(2\text{ pm}\) to \(3\text{ pm}\), the total duration is \(60\) minutes. If \(t\) minutes have passed, the remaining time to \(3\text{ pm}\) is \((60 – t)\) minutes.
According to the given condition:
\[60 – t = \frac{t^2}{4} – 3\] \[63 – t = \frac{t^2}{4} \implies 4(63 – t) = t^2 \implies 252 – 4t = t^2\] \[t^2 + 4t – 252 = 0\]Factoring the equation (\(18 \times (-14) = -252\)):
\[(t + 18)(t – 14) = 0 \implies t = 14 \quad \text{or} \quad t = -18\]Since time \(t\) must be positive, \(t = 14\). Hence, \(t = \mathbf{14\text{ minutes}}\).
