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CBSE Maths · Class 10

Chapter 4: Quadratic Equations

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Date: ______________
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Standard form, factorisation, completing the square, quadratic formula, nature of roots and word problems.

Questions

  1. Solve x23x10=0x^2 - 3x - 10 = 0.
  2. Discuss nature of roots of 2x26x+3=02x^2 - 6x + 3 = 0.
  3. Solve x+1/x=10/3x + 1/x = 10/3.
  4. Find kk so x2+2(k+1)x+k2=0x^2 + 2(k+1)x + k^2 = 0 has equal roots.
  5. Solve 3x2+10x+73=0\sqrt 3 x^2 + 10x + 7\sqrt 3 = 0.
  6. Solve x25x=0x^2 - 5x = 0.
  7. Product of two consecutive positive integers =306= 306. Find them.
  8. Sum of squares of two consecutive odd positive integers =290= 290. Find them.
  9. Roots of 2x2+x4=02x^2 + x - 4 = 0 by quadratic formula.
  10. A train travels 360360 km at uniform speed. If speed had been 55 km/h more, it would have taken 11 h less. Find its speed.

Extra Practice (Optional)

  • Solve 2x25x+32=0\sqrt 2 x^2 - 5x + 3\sqrt 2 = 0.
  • Discriminant of 3x243x+43x^2 - 4\sqrt 3 x + 4.
  • Two-digit number problem: sum digits 99, product 1818.
  • Age problem quadratic form.
  • Speed-time problem: boat's speed in still water.

Answer Key (For Teachers)

  1. Answer: 5,25, -2.
    Working: Split: 3=5+2-3 = -5 + 2; product 10-10.(x5)(x+2)(x-5)(x+2).
  2. Answer: Two distinct real.
    Working: D=3624=12>0D = 36 - 24 = 12 > 0.
  3. Answer: 33 or 1/31/3.
    Working: 3x210x+3=03x^2 - 10x + 3 = 0; factor (3x1)(x3)(3x-1)(x-3).
  4. Answer: k=1/2k = -1/2.
    Working: D=4(k+1)24k2=0D = 4(k+1)^2 - 4k^2 = 0; 2k+1=02k+1 = 0.
  5. Answer: 3,  7/3-\sqrt 3,\; -7/\sqrt 3.
    Working: Split middle: 10=7+310 = 7 + 3.Factor as (3x+7)(x+3)(\sqrt 3 x + 7)(x + \sqrt 3).
  6. Answer: 00 or 55.
    Working: x(x5)=0x(x-5) = 0.
  7. Answer: 17,1817, 18.
    Working: n(n+1)=306n(n+1) = 306; n=17n = 17.
  8. Answer: 11,1311, 13.
    Working: (2n1)2+(2n+1)2=2908n2+2=290n=6(2n-1)^2 + (2n+1)^2 = 290 \Rightarrow 8n^2+2=290 \Rightarrow n=6.
  9. Answer: 1±334\dfrac{-1\pm\sqrt{33}}{4}.
    Working: x=1±1+324=1±334x = \dfrac{-1\pm\sqrt{1+32}}{4} = \dfrac{-1\pm\sqrt{33}}{4}.
  10. Answer: v=40v = 40 km/h.
    Working: Let speed =v= v: 360v360v+5=1\dfrac{360}{v} - \dfrac{360}{v+5} = 1.360(v+5v)=v(v+5)v2+5v1800=0(v40)(v+45)=0360(v+5-v) = v(v+5) \Rightarrow v^2 + 5v - 1800 = 0 \Rightarrow (v-40)(v+45)=0.
Gyaan Maths · CBSE Class 10 · Chapter 4
Gyaan Maths — CBSE Class 6-10, built for board-toppers.
v1.1 · Feb 2026