Class 10 Maths – Quadratic Equations Practice Paper

Source: Original question

Concept Used: Recognising the standard form of a quadratic equation \(ax^2 + bx + c = 0\), \(a \neq 0\).

A quadratic equation has \(x^2\) as the highest power of the variable. In \(2x + 3 = 0\), the highest power of \(x\) is \(1\), so it is a linear equation, not a quadratic equation. All the other options contain an \(x^2\) term with a non-zero coefficient.

Source: Original question

Concept Used: Verifying whether a given value is a root by substitution (\(x = \alpha\) is a root of \(p(x) = 0\) if \(p(\alpha) = 0\)).

Substitute \(x = 2\) in each option:
(A) \(2^2 + 3(2) - 10 = 4 + 6 - 10 = 0\) ✓
(B) \(4 - 6 + 10 = 8\) ✗
(C) \(4 - 4 + 5 = 5\) ✗
(D) \(2(4) - 10 + 4 = 2\) ✗

Only option (A) gives zero, so \(x = 2\) is a root of \(x^2 + 3x - 10 = 0\).

Source: Original question

Concept Used: Solving a quadratic equation by factorisation.

\(x^2 - 7x + 10 = 0\)
Split the middle term: \(x^2 - 2x - 5x + 10 = 0\)
\(x(x - 2) - 5(x - 2) = 0 \Rightarrow (x - 2)(x - 5) = 0\)
Hence \(x = 2\) or \(x = 5\).

Source: CBSE Board Paper 2024 (Mathematics Standard, 30/2/1)

Concept Used: Nature of roots from the discriminant \(D = b^2 - 4ac\).

For \(x^2 + x + 1 = 0\): \(a = 1\), \(b = 1\), \(c = 1\).
\(D = (1)^2 - 4(1)(1) = 1 - 4 = -3\).

Since \(D < 0\), the equation has no real roots.

Source: CBSE Board Paper 2024 (Mathematics Standard, 30/1/1)

Concept Used: Discriminant condition for real and equal roots: \(D = 0\).

For \(ax^2 + bx + c = 0\), \(D = b^2 - 4ac\).
Real and equal roots \(\Rightarrow D = 0\):
\(b^2 - 4ac = 0 \Rightarrow b^2 = 4ac \Rightarrow ac = \frac{b^2}{4}\).

Source: CBSE Board Paper 2024 (Mathematics Standard, 30/5/1)

Concept Used: Sum and product of roots: sum \(= -\frac{b}{a}\), product \(= \frac{c}{a}\).

For \(5x^2 - 6x + 21 = 0\): \(a = 5\), \(b = -6\), \(c = 21\).
Sum of roots \(= -\frac{(-6)}{5} = \frac{6}{5}\).
Product of roots \(= \frac{21}{5}\).

Ratio of sum to product \(= \frac{6}{5} : \frac{21}{5} = 6 : 21 = 2 : 7\).

Source: CBSE Board Paper 2024 (Mathematics Standard, 30/4/1)

Concept Used: Discriminant \(D = b^2 - 4ac\) and solving for an unknown constant.

For \(3x^2 - 2x + c = 0\): \(a = 3\), \(b = -2\), \(c = c\).
\(D = (-2)^2 - 4(3)(c) = 4 - 12c\).

Given \(D = 16\): \(4 - 12c = 16 \Rightarrow 12c = -12 \Rightarrow c = -1\).

Source: CBSE Board Paper 2025 (Mathematics Standard, 30/4/2)

Concept Used: Equal roots (\(D = 0\)) combined with the sign of the roots.

For \(ax^2 + x + a = 0\): \(b = 1\), so \(D = 1 - 4a^2\).
Equal roots \(\Rightarrow D = 0 \Rightarrow 1 - 4a^2 = 0 \Rightarrow a = \pm \frac{1}{2}\).

For equal roots, the common root is \(x = -\frac{b}{2a} = -\frac{1}{2a}\).
If \(a = \frac{1}{2}\): \(x = -1\) (negative).
If \(a = -\frac{1}{2}\): \(x = 1\) (positive). ✓

Hence \(a = -\frac{1}{2}\) gives equal and positive roots.

Source: Original question

Concept Used: Discriminant condition for two distinct real roots.

Assertion: For \(x^2 - 4x + 3 = 0\): \(a = 1\), \(b = -4\), \(c = 3\).
\(D = (-4)^2 - 4(1)(3) = 16 - 12 = 4 > 0\), so the equation has two distinct real roots. The Assertion is true.

Reason: The statement that \(D > 0\) implies two distinct real roots is a standard, correct rule, and it is exactly the rule that establishes the Assertion. The Reason is true and correctly explains A.

Source: Original question

Concept Used: Nature of roots; distinguishing a general fact from the specific reason for equality.

Assertion: For \(x^2 - 6x + 9 = 0\): \(a = 1\), \(b = -6\), \(c = 9\).
\(D = 36 - 36 = 0\), so the equation has two real and equal roots. The Assertion is true.

Reason: A quadratic equation certainly has at most two real roots — this is a true general fact. However, it does not state why this particular equation has equal roots (the specific reason is \(D = 0\)). Hence R is true but not the correct explanation of A.

Source: CBSE Board Paper 2025 (Mathematics Standard, 30/3/1)

Concept Used: Real and equal roots \(\Rightarrow D = 0\).

For \(4x^2 + kx + 1 = 0\): \(a = 4\), \(b = k\), \(c = 1\).
\(D = k^2 - 4(4)(1) = k^2 - 16\).
Equal roots \(\Rightarrow D = 0 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4\).

Verification: For \(k = 4\), \(4x^2 + 4x + 1 = (2x + 1)^2\). For \(k = -4\), \(4x^2 - 4x + 1 = (2x - 1)^2\). Both are perfect squares (equal roots).

Source: Original question

Concept Used: Solving by splitting the middle term (factorisation).

\(2x^2 - 5x + 3 = 0\)
Find two numbers whose product is \(2 \times 3 = 6\) and whose sum is \(-5\): they are \(-2\) and \(-3\).
\(2x^2 - 2x - 3x + 3 = 0\)
\(2x(x - 1) - 3(x - 1) = 0 \Rightarrow (x - 1)(2x - 3) = 0\)

Hence \(x = 1\) or \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\).

Verification: \(x = 1\): \(2 - 5 + 3 = 0\) ✓   \(x = \frac{3}{2}\): \(2(\frac{9}{4}) - \frac{15}{2} + 3 = \frac{9}{2} - \frac{15}{2} + \frac{6}{2} = 0\) ✓

Source: Original question

Concept Used: Forming a quadratic equation from a stated condition on numbers.

Let the two consecutive positive integers be \(x\) and \(x + 1\).
\(x(x + 1) = 240 \Rightarrow x^2 + x - 240 = 0\)

\(x^2 + x - 240 = 0 \Rightarrow x^2 + 16x - 15x - 240 = 0\)
\(x(x + 16) - 15(x + 16) = 0 \Rightarrow (x + 16)(x - 15) = 0\)
\(x = -16\) (rejected, since the integer must be positive) or \(x = 15\).

The integers are \(15\) and \(16\). Check: \(15 \times 16 = 240\) ✓

Source: Original question

Concept Used: Discriminant and nature of roots.

For \(3x^2 - 2x - 1 = 0\): \(a = 3\), \(b = -2\), \(c = -1\).
\(D = b^2 - 4ac = (-2)^2 - 4(3)(-1) = 4 + 12 = 16\).

Since \(D = 16 > 0\) is a perfect square, the equation has two distinct real and rational roots. (Indeed \(x = \frac{2 \pm 4}{6}\), giving \(x = 1\) and \(x = -\frac{1}{3}\).)

Source: Original question

Concept Used: Forming and solving a quadratic equation from a geometry (area) condition.

Let the breadth of the garden be \(x\) m. Its length is \(2x + 4\) m.
Area \(= x(2x + 4) = 160 \Rightarrow 2x^2 + 4x - 160 = 0\)
Dividing by 2: \(x^2 + 2x - 80 = 0\)
\((x + 10)(x - 8) = 0 \Rightarrow x = -10\) or \(x = 8\)

Reject \(x = -10\) (breadth cannot be negative). Hence breadth \(= 8\) m and length \(= 2(8) + 4 = 20\) m.

Verification: Area \(= 20 \times 8 = 160\) m² ✓

Source: Original question

Concept Used: Age problems — forming and solving a quadratic equation.

Let the daughter's present age be \(x\) years. Then the woman's present age is \(x^2\) years.
One year ago: daughter \(x - 1\), woman \(x^2 - 1\).
\(x^2 - 1 = 7(x - 1) \Rightarrow x^2 - 7x + 6 = 0\)
\((x - 1)(x - 6) = 0 \Rightarrow x = 1\) or \(x = 6\)

\(x = 1\) is rejected because one year ago the daughter's age would be \(0\), for which "seven times as old" cannot hold. Hence \(x = 6\).

Daughter \(= 6\) years, woman \(= 6^2 = 36\) years. Check: One year ago \(35 = 7 \times 5\) ✓

Source: CBSE Board Paper 2024 (Mathematics Standard, 30/5/1)

Concept Used: Equal roots condition \(D = 0\) applied to an equation with a parameter.

For \((k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0\):
\(a = k + 1\), \(b = -6(k + 1)\), \(c = 3(k + 9)\).

\(D = [-6(k + 1)]^2 - 4(k + 1) \cdot 3(k + 9)\)
\(= 36(k + 1)^2 - 12(k + 1)(k + 9)\)
\(= 12(k + 1)[3(k + 1) - (k + 9)]\)
\(= 12(k + 1)(2k - 6) = 24(k + 1)(k - 3)\)

Equal roots \(\Rightarrow D = 0 \Rightarrow k + 1 = 0\) or \(k - 3 = 0\).
\(k = -1\) is rejected (given \(k \neq -1\)), so \(k = 3\).

Verification: With \(k = 3\): \(4x^2 - 24x + 36 = 4(x - 3)^2\), giving equal roots \(3, 3\).

Source: Original question

Concept Used: Speed–distance–time application; forming a quadratic equation.

Let the speed of the stream be \(x\) km/h.
Upstream speed \(= 18 - x\), downstream speed \(= 18 + x\).

Time upstream \(= \frac{24}{18 - x}\), time downstream \(= \frac{24}{18 + x}\).
\(\frac{24}{18 - x} - \frac{24}{18 + x} = 1\)
\(24\left(\frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)}\right) = 1 \Rightarrow \frac{24(2x)}{324 - x^2} = 1\)
\(48x = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0\)

\(x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm 60}{2}\)
\(x = 6\) or \(x = -54\). Reject \(x = -54\) (speed cannot be negative). Hence \(x = 6\).

Verification: Upstream \(24/12 = 2\) h, downstream \(24/24 = 1\) h, difference 1 h ✓

Source: Original question

Concept Used: Speed–distance–time application (reduced time when speed increases).

Let the speed of the train be \(x\) km/h.
Time at speed \(x\): \(\frac{300}{x}\) h. Time at speed \(x + 5\): \(\frac{300}{x + 5}\) h.

\(\frac{300}{x} - \frac{300}{x + 5} = 2\)
\(300\left(\frac{5}{x(x + 5)}\right) = 2 \Rightarrow 1500 = 2x(x + 5)\)
\(x^2 + 5x - 750 = 0 \Rightarrow (x + 30)(x - 25) = 0\)
\(x = -30\) (rejected) or \(x = 25\).

Verification: \(300/25 = 12\) h and \(300/30 = 10\) h; difference \(2\) h ✓

Source: Original question

Concept Used: Reducing an equation to standard quadratic form and interpreting its discriminant.

\(x + \frac{1}{x} = 2\), \(x \neq 0\).
Multiplying throughout by \(x\): \(x^2 + 1 = 2x \Rightarrow x^2 - 2x + 1 = 0\).

\((x - 1)^2 = 0 \Rightarrow x = 1\).

Why only one distinct solution: For \(x^2 - 2x + 1 = 0\), \(a = 1\), \(b = -2\), \(c = 1\), so \(D = (-2)^2 - 4(1)(1) = 0\). A zero discriminant means the two roots are real and equal (a repeated root), so the equation has only one distinct solution, \(x = 1\).

Verification: \(1 + \frac{1}{1} = 2\) ✓

Source: Original question

Concept Used: Two-digit-number problem — forming and solving a quadratic equation.

Let the tens digit be \(x\) and the units digit be \(y\). The number is \(10x + y\).
Product condition: \(xy = 21\).  â€¦(1)
Interchange condition: \(10x + y - 36 = 10y + x \Rightarrow 9x - 9y = 36 \Rightarrow x - y = 4\).  â€¦(2)

From (1), \(y = \frac{21}{x}\). Putting this in (2):
\(x - \frac{21}{x} = 4 \Rightarrow x^2 - 4x - 21 = 0 \Rightarrow (x - 7)(x + 3) = 0\)
\(x = 7\) (rejecting \(x = -3\), since a digit cannot be negative). Then \(y = \frac{21}{7} = 3\).

The number is \(10(7) + 3 = 73\). Check: \(73 - 36 = 37\) (digits interchanged) and \(7 \times 3 = 21\) ✓

Source: CBSE Board Paper 2025 (Mathematics Standard, 30/1/1)

Concept Used: Speed–distance–time application (more time taken when speed reduces).

Let the uniform speed be \(x\) km/h.
Time at speed \(x\): \(\frac{480}{x}\) h. Time at speed \(x - 8\): \(\frac{480}{x - 8}\) h.

\(\frac{480}{x - 8} - \frac{480}{x} = 3\)
\(480\left(\frac{x - (x - 8)}{x(x - 8)}\right) = 3 \Rightarrow \frac{480 \times 8}{x(x - 8)} = 3\)
\(3840 = 3x(x - 8) \Rightarrow x^2 - 8x - 1280 = 0 \Rightarrow (x - 40)(x + 32) = 0\)
\(x = 40\) (rejecting \(x = -32\)).

Verification: \(480/40 = 12\) h and \(480/32 = 15\) h; difference \(3\) h ✓

Source: Original question — Case Study

Concept Used: Interpreting a real-life situation as a quadratic equation and justifying the valid root.

(a) Breadth \(= x\) m, length \(= x + 4\) m.
Area \(= x(x + 4) = 96 \Rightarrow x^2 + 4x - 96 = 0\).

(b) \(x^2 + 4x - 96 = 0 \Rightarrow x^2 + 12x - 8x - 96 = 0\)
\(x(x + 12) - 8(x + 12) = 0 \Rightarrow (x + 12)(x - 8) = 0\)
\(x = -12\) or \(x = 8\).

The root \(x = -12\) is rejected because the breadth (a length) cannot be negative. Hence breadth \(= 8\) m and length \(= 12\) m.

Verification: \(8 \times 12 = 96\) m² and \(12 - 8 = 4\) m ✓