Class 10 Quadratic Equations Mock Test (CBSE)

Topics in Quadratic Equations

• Quadratic equations
• Solution by factorisation
• Quadratic formula
• Nature of roots and the discriminant

Quadratic Equations practice questions with answers

1. Q1Multiple choice·1 mark
   If one root of the quadratic equation $x^2 - 8x + m = 0$ exceeds the other by $2$, then the value of $m$ is:

   1. (a)$15$
   2. (b)$16$
   3. (c)$12$
   4. (d)$14$
   Show answer and step marks

   1. Correct option: (a) $15$. Let the roots be $\alpha$ and $\alpha + 2$. Sum of roots = $2\alpha + 2 = 8 \implies 2\alpha = 6 \implies \alpha = 3$. The roots are $3$ and $5$. Product of roots $m = 3 \times 5 = 15$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Which of the following quadratic equations has $x = 2$ as one of its roots?

   1. (a)$x^2 - 4x + 5 = 0$
   2. (b)$x^2 + 3x - 12 = 0$
   3. (c)$2x^2 - 7x + 6 = 0$
   4. (d)$3x^2 - 6x - 2 = 0$
   Show answer and step marks

   1. Correct option: (c) $2x^2 - 7x + 6 = 0$. Substitute $x = 2$ into the equations. For $2x^2 - 7x + 6 = 0$, we get $2(2)^2 - 7(2) + 6 = 8 - 14 + 6 = 0$.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   The roots of the quadratic equation $x^2 - 3x - 10 = 0$ are:

   1. (a)$5, -2$
   2. (b)$-5, 2$
   3. (c)$5, 2$
   4. (d)$-5, -2$
   Show answer and step marks

   1. Correct option: (a) $5, -2$. Factorising $x^2 - 3x - 10 = 0$ gives $(x - 5)(x + 2) = 0$, which yields roots $5$ and $-2$.+1

   Answer: (a)
4. Q4Short answer·3 marks
   Find the values of $k$ for which the quadratic equation $(k + 4)x^2 + (k + 1)x + 1 = 0$ has equal roots. Also, find the roots of the equation for each value of $k$.
   Show answer and step marks

   1. For equal roots $D = b^2 - 4ac = 0$ with $a = k + 4$, $b = k + 1$, $c = 1$: $(k + 1)^2 - 4(k + 4) = 0$.+1
   2. Expand: $k^2 + 2k + 1 - 4k - 16 = 0 \implies k^2 - 2k - 15 = 0$.+0.5
   3. Factorise: $(k - 5)(k + 3) = 0$, so $k = 5$ or $k = -3$.+0.5
   4. For $k = 5$: $9x^2 + 6x + 1 = 0 \implies (3x + 1)^2 = 0$, so $x = -\frac{1}{3}, -\frac{1}{3}$.+0.5
   5. For $k = -3$: $x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0$, so $x = 1, 1$.+0.5

   Answer: $k = 5$, roots: $x = -\frac{1}{3}, -\frac{1}{3}$; $k = -3$, roots: $x = 1, 1$
5. Q5Short answer·3 marks
   The length of a rectangular garden is $4$ m more than its breadth. The area of the garden is $60$ m². Let the breadth of the garden be $x$ metres. (a) Form the quadratic equation in $x$ from the given area. (b) Solve the quadratic equation formed in part (a) by factorisation. (c) Explain why one of the roots of the equation cannot be the breadth of the garden.
   Show answer and step marks

   1. Length $= (x + 4)$ m, so $x(x + 4) = 60$, which gives $x^2 + 4x - 60 = 0$.+1
   2. Factorising: $x^2 + 4x - 60 = (x + 10)(x - 6) = 0$, so $x = 6$ or $x = -10$.+1
   3. $x = -10$ is rejected because a breadth (a length) cannot be negative; the breadth is $6$ m.+1

   Answer: (a) $x^2 + 4x - 60 = 0$; (b) $x = 6$ or $x = -10$; (c) $x = -10$ is rejected since a length cannot be negative
6. Q6Long answer·5 marks
   The quadratic equation $x^2 + kx + 4 = 0$, where $k$ is a real number, is given. (a) Write the discriminant of the equation in terms of $k$. [1] (b) Find the values of $k$ for which the equation has equal real roots. [1] (c) Find the range of values of $k$ for which the equation has two distinct real roots. [2] (d) Find the nature of the roots of $x^2 + 5x + 4 = 0$ using its discriminant. [1]
   Show answer and step marks

   1. (a) Compare with $ax^2 + bx + c = 0$: $a = 1$, $b = k$, $c = 4$. Discriminant $D = b^2 - 4ac = k^2 - 4(1)(4) = k^2 - 16$.+1
   2. (b) Equal real roots need $D = 0$, so $k^2 - 16 = 0$.+0.5
   3. $k = 4$ or $k = -4$.+0.5
   4. (c) Two distinct real roots need $D > 0$, so $k^2 - 16 > 0$.+1
   5. $(k-4)(k+4) > 0$ gives $k < -4$ or $k > 4$.+1
   6. (d) For $x^2 + 5x + 4 = 0$: $D = 5^2 - 4(1)(4) = 25 - 16 = 9$.+0.5
   7. Since $D = 9 > 0$, the roots are real and distinct (unequal).+0.5

   Answer: (a) $D = k^2 - 16$; (b) $k = 4$ or $k = -4$; (c) $k < -4$ or $k > 4$; (d) $D = 9 > 0$, so the roots are real and distinct.

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