Quadratic Equations — Class 10 CBSE Mathematics MCQs with Solutions

Solution: Expanding $(x+1)^2 = 2(x-3)$ gives $x^2 + 2x + 1 = 2x - 6$, i.e. $x^2 + 7 = 0$, which is a polynomial equation of degree 2, so it is quadratic. The first option contains $\sqrt{x}$, so it is not a polynomial equation. The third simplifies to $-3x + 1 = 0$ (linear), and the fourth becomes $x^4 - 5x^2 + 1 = 0$ (degree 4).

Solution: If $a = 0$, the term $ax^2$ vanishes and the equation becomes $bx + c = 0$, which is linear, not quadratic. So the only condition needed is $a \neq 0$. The values of $b$ and $c$ may be zero, and $a$ may be negative, e.g. $-2x^2 + 5 = 0$ is a valid quadratic equation.

Solution: A value is a root if it satisfies the equation. Substituting $x = 2$: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$, so $x = 2$ is a root. Checking the others: $x = 1$ gives $2$, $x = 4$ gives $2$, and $x = 5$ gives $6$ — none of them is zero.

Solution: Splitting the middle term: $x^2 - 5x + 2x - 10 = 0$ gives $x(x-5) + 2(x-5) = 0$, i.e. $(x-5)(x+2) = 0$. So $x = 5$ or $x = -2$. Check: $25 - 15 - 10 = 0$ and $4 + 6 - 10 = 0$.

Solution: Taking $x$ common: $x(x - 7) = 0$, so $x = 0$ or $x = 7$. A common mistake is to divide both sides by $x$, which loses the root $x = 0$ — a quadratic equation always has two roots (counting repetition).

Solution: Using the difference of squares, $x^2 - 49 = (x - 7)(x + 7) = 0$, so $x = 7$ or $x = -7$. Remember that taking the square root of both sides gives both the positive and the negative root — writing only $x = 7$ misses a solution.

Solution: The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, obtained by completing the square on $ax^2 + bx + c = 0$. Watch the three common slips shown in the wrong options: forgetting the minus sign on $b$, writing $+4ac$ instead of $-4ac$ under the root, and dividing by $a$ instead of $2a$.

Solution: Here $a = 1$, $b = 4$, $c = 3$, so $b^2 - 4ac = 16 - 12 = 4$. Then $x = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2}$, giving $x = -1$ and $x = -3$. Check: $(-1)^2 + 4(-1) + 3 = 0$ and $(-3)^2 + 4(-3) + 3 = 0$.

Solution: The discriminant is $D = b^2 - 4ac$ with $a = 2$, $b = -4$, $c = 3$. So $D = (-4)^2 - 4(2)(3) = 16 - 24 = -8$. Since $D < 0$, this equation has no real roots; writing $16 + 24 = 40$ is the common sign error.

Solution: When $D = b^2 - 4ac > 0$, the square root $\sqrt{D}$ is a positive real number, so the formula $x = \frac{-b \pm \sqrt{D}}{2a}$ gives two different real values. Equal roots occur only when $D = 0$, and no real roots when $D < 0$. A quadratic equation can never have exactly one real root unless the two roots coincide (the $D = 0$ case, which counts as two equal roots).

Solution: Time at usual speed is $\frac{480}{x}$ hours and at the reduced speed it is $\frac{480}{x-8}$ hours, and the slower journey takes 3 hours more: $\frac{480}{x-8} - \frac{480}{x} = 3$. Multiplying through by $x(x-8)$: $480x - 480(x-8) = 3x(x-8)$, i.e. $3840 = 3x^2 - 24x$. Dividing by 3 gives $x^2 - 8x - 1280 = 0$.

Solution: Let the larger number be $x$. The square of the smaller number is $8x$, so $x^2 - 8x = 180$, i.e. $x^2 - 8x - 180 = 0$. Factorising: $(x - 18)(x + 10) = 0$, so $x = 18$ (rejecting $-10$ as the numbers are positive). Then the smaller number is $\sqrt{8 \times 18} = \sqrt{144} = 12$; check: $18^2 - 12^2 = 324 - 144 = 180$.

Solution: Let the number of articles be $x$; then the cost of each article is $₹(2x + 3)$. Total cost: $x(2x + 3) = 90$, i.e. $2x^2 + 3x - 90 = 0$. Factorising: $2x^2 + 15x - 12x - 90 = (2x + 15)(x - 6) = 0$, so $x = 6$ (rejecting the negative root). Check: each article costs ₹15 and $6 \times 15 = 90$.

Solution: Combining the fractions: $\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{-11}{(x+4)(x-7)} = \frac{11}{30}$, so $(x+4)(x-7) = -30$. Expanding: $x^2 - 3x - 28 = -30$, i.e. $x^2 - 3x + 2 = 0$, which factorises as $(x-1)(x-2) = 0$. So $x = 1$ or $x = 2$; check for $x = 1$: $\frac{1}{5} + \frac{1}{6} = \frac{11}{30}$.

Solution: Let the stream's speed be $x$ km/h; upstream speed is $(18 - x)$ and downstream is $(18 + x)$. The condition gives $\frac{24}{18-x} - \frac{24}{18+x} = 1$, which simplifies to $48x = 324 - x^2$, i.e. $x^2 + 48x - 324 = 0$. By the quadratic formula, $x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm 60}{2}$, so $x = 6$ km/h (rejecting $-54$).

Solution: Let the breadth be $b$ m; then $b(b + 3) = 180$, i.e. $b^2 + 3b - 180 = 0$. The discriminant is $9 + 720 = 729$ and $\sqrt{729} = 27$, so $b = \frac{-3 + 27}{2} = 12$ m (rejecting the negative root). The length is 15 m, so the perimeter is $2(12 + 15) = 54$ m — note the question asks for the perimeter, not the dimensions.

Solution: Let the smaller tap take $x$ hours; the larger takes $(x - 10)$ hours, and together they fill $\frac{8}{75}$ of the tank per hour since $9\frac{3}{8} = \frac{75}{8}$. So $\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}$, which gives $8x^2 - 230x + 750 = 0$, i.e. $4x^2 - 115x + 375 = 0$. The quadratic formula gives $x = \frac{115 \pm \sqrt{13225 - 6000}}{8} = \frac{115 \pm 85}{8}$, so $x = 25$ (rejecting $3.75$, which would make $x - 10$ negative). Check: $\frac{1}{25} + \frac{1}{15} = \frac{8}{75}$.

Solution: For equal roots, $D = (-2k)^2 - 4(k)(6) = 4k^2 - 24k = 0$, so $4k(k - 6) = 0$, giving $k = 0$ or $k = 6$. But if $k = 0$ the equation becomes $6 = 0$, which is not a quadratic equation at all, so $k = 0$ must be rejected. Hence $k = 6$; check: $6x^2 - 12x + 6 = 6(x-1)^2$, which has the repeated root $x = 1$.

Solution: For $x^2 - 4x + 3\sqrt{2} = 0$, the discriminant is $16 - 12\sqrt{2} \approx 16 - 16.97 < 0$, so it has no real roots. In the first two options the constant term is negative, making $D = 16 + 12\sqrt{2} > 0$ (real roots exist), and for $3x^2 + 4\sqrt{3}x + 4 = 0$ we get $D = 48 - 48 = 0$ (real equal roots). The key comparison is $12\sqrt{2} > 16$ since $\sqrt{2} \approx 1.414$.

Solution: For real roots, the discriminant must be non-negative: $k^2 - 4(2)(2) \geq 0$, i.e. $k^2 \geq 16$. This means $|k| \geq 4$, so $k \leq -4$ or $k \geq 4$ — remember that $k^2 \geq 16$ includes large negative values of $k$ too. For $-4 < k < 4$ the discriminant is negative and there are no real roots, while $k = \pm 4$ gives exactly equal roots (a special case, not the whole answer).

Solution: Since $x = \frac{1}{2}$ is a root, it satisfies the equation: $2\left(\frac{1}{4}\right) + \frac{k}{2} - 3 = 0$. This gives $\frac{1}{2} + \frac{k}{2} = 3$, so $\frac{k}{2} = \frac{5}{2}$ and $k = 5$. Check: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$, which is indeed zero at $x = \frac{1}{2}$.

Solution: Expanding the left side: $(x-3)(2x+1) = 2x^2 - 5x - 3$. The right side is $x^2 + 5x$. Bringing everything to one side: $2x^2 - 5x - 3 - x^2 - 5x = 0$, which simplifies to $x^2 - 10x - 3 = 0$. Since the degree is 2, the given equation is quadratic.

Solution: We need two numbers whose product is $6 \times (-2) = -12$ and sum is $-1$: they are $-4$ and $3$. So $6x^2 - 4x + 3x - 2 = 0$ gives $2x(3x - 2) + 1(3x - 2) = 0$, i.e. $(3x - 2)(2x + 1) = 0$. Hence $x = \frac{2}{3}$ or $x = -\frac{1}{2}$.

Solution: Split the middle term: $\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$ (since $2 \times 5 = 10 = \sqrt{2} \times 5\sqrt{2}$). Grouping: $\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$, so $(x + \sqrt{2})(\sqrt{2}x + 5) = 0$. Hence $x = -\sqrt{2}$ or $x = -\frac{5}{\sqrt{2}}$; both roots are negative because all coefficients are positive.

Solution: Here $a = 2$, $b = -7$, $c = 3$, so the discriminant is $(-7)^2 - 4(2)(3) = 49 - 24 = 25$. Then $x = \frac{7 \pm 5}{4}$, giving $x = 3$ and $x = \frac{1}{2}$. Note that $-b = -(-7) = 7$; keeping the sign wrong would give the negatives of the roots.

Solution: The discriminant is $16 - 4 = 12$, so $\sqrt{12} = 2\sqrt{3}$. By the quadratic formula, $x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ — remember to divide both terms of the numerator by 2. These are irrational roots; check: $(2+\sqrt{3})^2 - 4(2+\sqrt{3}) + 1 = 7 + 4\sqrt{3} - 8 - 4\sqrt{3} + 1 = 0$.

Solution: Let the integers be $x$ and $x + 1$: $x^2 + (x+1)^2 = 365$ gives $2x^2 + 2x - 364 = 0$, i.e. $x^2 + x - 182 = 0$. By the quadratic formula, $x = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm 27}{2}$, so $x = 13$ (rejecting $-14$). The integers are 13 and 14; check: $169 + 196 = 365$.

Solution: For equal roots, the discriminant must be zero: $k^2 - 4(2)(3) = 0$, so $k^2 = 24$ and $k = \pm\sqrt{24} = \pm 2\sqrt{6}$. Both values are valid — dropping the negative value (writing $2\sqrt{6}$ only) loses half the answer.

Solution: The discriminant is $D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$. Since $D = 0$, the equation has two equal real roots, each equal to $\frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}}$. Note that $(-4\sqrt{3})^2 = 16 \times 3 = 48$, not $-48$.

Solution: The discriminant is $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$. Since $D > 0$, the equation has two distinct real roots (in fact $x = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$). The discriminant alone decides the nature of the roots — no need to solve the equation fully.