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Chapter 4: Quadratic Equations

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This page provides comprehensive Chapter 4: Quadratic Equations - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Quadratic Equations. Practice for CBSE Board Exams.

Standard Level Worksheet

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  1. Question 1: Find the roots of the quadratic equation $2x^2 - x + \frac{1}{8} = 0$ by factorization.
    Solution:
    Step 1: Multiply by 8 to remove fraction: $16x^2 - 8x + 1 = 0$.
    Step 2: Factorize: $(4x)^2 - 2(4x)(1) + 1^2 = 0 \Rightarrow (4x-1)^2 = 0$.
    Step 3: $4x - 1 = 0 \Rightarrow x = 1/4$.
    Answer: Roots are $1/4, 1/4$.
  2. Question 2: Find the roots of the quadratic equation $3x^2 - 5x + 2 = 0$ using the quadratic formula.
    Solution:
    Step 1: $a=3, b=-5, c=2$. Discriminant $D = b^2 - 4ac = (-5)^2 - 4(3)(2) = 25 - 24 = 1$.
    Step 2: $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{5 \pm 1}{6}$.
    Step 3: $x = \frac{6}{6} = 1$ or $x = \frac{4}{6} = \frac{2}{3}$.
    Answer: Roots are $1, 2/3$.
  3. Question 3: Find the value of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two real equal roots.
    Solution:
    Step 1: For equal roots, $D = 0 \Rightarrow b^2 - 4ac = 0$.
    Step 2: $k^2 - 4(2)(3) = 0 \Rightarrow k^2 - 24 = 0$.
    Step 3: $k^2 = 24 \Rightarrow k = \pm\sqrt{24} = \pm 2\sqrt{6}$.
    Answer: $k = \pm 2\sqrt{6}$.
  4. Question 4: Find the nature of the roots of the quadratic equation $2x^2 - 6x + 3 = 0$.
    Solution:
    Step 1: Calculate Discriminant $D = b^2 - 4ac$.
    Step 2: $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$.
    Step 3: Since $D > 0$, the roots are real and distinct.
    Answer: Real and distinct roots.
  5. Question 5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
    Solution:
    Step 1: Let base $= x$ cm. Altitude $= x - 7$ cm.
    Step 2: By Pythagoras theorem: $x^2 + (x-7)^2 = 13^2$.
    Step 3: $x^2 + x^2 - 14x + 49 = 169 \Rightarrow 2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0$.
    Step 4: $(x-12)(x+5) = 0 \Rightarrow x = 12$ (since side cannot be negative).
    Answer: Base = 12 cm, Altitude = 5 cm.
  6. Question 6: Find two consecutive odd positive integers, sum of whose squares is 290.
    Solution:
    Step 1: Let integers be $x$ and $x+2$.
    Step 2: $x^2 + (x+2)^2 = 290 \Rightarrow x^2 + x^2 + 4x + 4 = 290$.
    Step 3: $2x^2 + 4x - 286 = 0 \Rightarrow x^2 + 2x - 143 = 0$.
    Step 4: $(x+13)(x-11) = 0 \Rightarrow x = 11$ (since positive).
    Answer: Integers are 11 and 13.
  7. Question 7: Solve for $x$: $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq -4, 7$.
    Solution:
    Step 1: Take LCM: $\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}$.
    Step 2: $\frac{-11}{x^2 - 3x - 28} = \frac{11}{30} \Rightarrow -1 = \frac{x^2 - 3x - 28}{30}$.
    Step 3: $x^2 - 3x - 28 = -30 \Rightarrow x^2 - 3x + 2 = 0$.
    Step 4: $(x-2)(x-1) = 0 \Rightarrow x = 1, 2$.
    Answer: $x = 1, 2$.
  8. Question 8: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
    Solution:
    Step 1: Let speed $= x$ km/h. Time $= 360/x$. New speed $= x+5$. New time $= 360/(x+5)$.
    Step 2: $\frac{360}{x} - \frac{360}{x+5} = 1$.
    Step 3: $360(x+5-x) = x(x+5) \Rightarrow 1800 = x^2 + 5x$.
    Step 4: $x^2 + 5x - 1800 = 0 \Rightarrow (x+45)(x-40) = 0 \Rightarrow x = 40$ (speed > 0).
    Answer: Speed is 40 km/h.
  9. Question 9: If $-5$ is a root of the quadratic equation $2x^2 + px - 15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots, find the value of $k$.
    Solution:
    Step 1: Put $x=-5$ in eq 1: $2(-5)^2 + p(-5) - 15 = 0 \Rightarrow 50 - 5p - 15 = 0 \Rightarrow 5p = 35 \Rightarrow p = 7$.
    Step 2: Eq 2 becomes $7(x^2 + x) + k = 0 \Rightarrow 7x^2 + 7x + k = 0$.
    Step 3: For equal roots, $D = 0 \Rightarrow 7^2 - 4(7)(k) = 0 \Rightarrow 49 - 28k = 0 \Rightarrow k = 49/28 = 7/4$.
    Answer: $k = 7/4$.
  10. Question 10: A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find its length and breadth.
    Solution:
    Step 1: Let length $= x$, breadth $= x-3$. Area Rect $= x(x-3)$.
    Step 2: Area Triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(x-3)(12) = 6(x-3)$.
    Step 3: $x(x-3) = 6(x-3) + 4 \Rightarrow x^2 - 3x = 6x - 18 + 4 \Rightarrow x^2 - 9x + 14 = 0$.
    Step 4: $(x-7)(x-2) = 0 \Rightarrow x = 7$ or $x = 2$. If $x=2$, breadth is negative. So $x=7$.
    Answer: Length = 7 m, Breadth = 4 m.
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