Mastersheet Solutions: Quadratic Equations | Class 10 CBSE

Topic 1 Solutions: MCQ / 1-Mark

Ans: (b) $2x^2 + 3 = 0$. A quadratic equation has degree exactly 2. Option (a) is cubic; (c) reduces to $x^2-2x+1=0$ ✓ but also valid; (d) $x^2+x+1=x^2-7x+10$ simplifies to $8x-9=0$ (linear). Option (b) is a standard quadratic. Also (c): $x\cdot x + 1 = x^2-7x+10 \implies 8x=9$, linear. Best answer: (b).

Ans: (a) 5 and –2. $x^2-3x-10=(x-5)(x+2)=0 \implies x=5$ or $x=-2$.

Ans: (a) –8. $D = b^2-4ac = (-4)^2 - 4(2)(3) = 16-24 = \mathbf{-8}$.

Ans: (b) Equal and real. When $D=0$: roots $= \dfrac{-b \pm 0}{2a} = \dfrac{-b}{2a}$ — one unique real value (counted as two equal roots).

Ans: (a) 1. $D=0$: $4-4k=0 \implies k=1$.

Ans: (b) 2 and 2. $2x^2-8x+8 = 2(x-2)^2=0 \implies x=2$ (equal roots).

Ans: (d) Both (a) and (c). $p(1) = 1+1-2=0$ ✓ for (a); $2(1)-3+1=0$ ✓ for (c).

Ans: (c) 3. $3\!\left(\dfrac{1}{9}\right)-\dfrac{10}{3}+k=0 \implies \dfrac{1}{3}-\dfrac{10}{3}+k=0 \implies k=3$.

Ans: (c) No real roots. $(x-1)^2+x^2 = x^2-2x+1+x^2 = 2x^2-2x+1=0$. $D=4-8=-4<0$. No real roots.

10.

Ans: (b) Equal. $D = 36-36=0$. So roots are real and equal: $x=\dfrac{1}{3}$ (both roots).

11.

Ans: (a) 0. $p\!\left(\dfrac{2}{3}\right) = 3\cdot\dfrac{4}{9}-5\cdot\dfrac{2}{3}+2 = \dfrac{4}{3}-\dfrac{10}{3}+\dfrac{6}{3} = 0$ ✓.

12.

Ans: (a) $\dfrac{3}{5}$. Sum of roots $= -\dfrac{b}{a} = -\dfrac{-3}{5} = \dfrac{3}{5}$.

Topic 2 Solutions: Standard Form Checks

13.

Ans:

$(x+1)^2=2(x-3) \implies x^2+2x+1=2x-6 \implies x^2+7=0$. Degree 2, $a\neq0$. Yes, quadratic.
$x^2-2x=-2(3-x)=-6+2x \implies x^2-4x+6=0$. Degree 2. Yes, quadratic.
$(x-2)(x+1)=(x-1)(x+3) \implies x^2-x-2=x^2+2x-3 \implies -3x+1=0$. Degree 1. Not quadratic.
$(x+2)^3=2x(x^2-1) \implies x^3+6x^2+12x+8=2x^3-2x \implies x^3-6x^2-14x-8=0$. Degree 3. Not quadratic.

14.

Ans:

Let breadth $= b$. Length $= 2b+1$. Area: $b(2b+1)=528 \implies 2b^2+b-528=0$.
Let integers be $n$ and $n+1$. $n(n+1)=306 \implies n^2+n-306=0$.

15.

Ans: $p(-3)=9-18+9=0$. Yes, $x=-3$ is a solution.

Topic 3 Solutions: Factorisation Method

16.

Ans: $2x^2+x-6=2x^2+4x-3x-6=2x(x+2)-3(x+2)=(2x-3)(x+2)=0$. Roots: $x=\dfrac{3}{2}$ and $x=-2$.

17.

Ans: $\sqrt{2}\,x^2+7x+5\sqrt{2}=\sqrt{2}\,x^2+5x+2x+5\sqrt{2}=x(\sqrt{2}\,x+5)+\sqrt{2}(\sqrt{2}\,x+5)=(x+\sqrt{2})(\sqrt{2}\,x+5)=0$. Roots: $x=-\sqrt{2}$ and $x=-\dfrac{5}{\sqrt{2}}=-\dfrac{5\sqrt{2}}{2}$.

18.

Ans: $100x^2-20x+1=(10x-1)^2=0 \implies x=\dfrac{1}{10}$ (equal roots). Root: $x=\dfrac{1}{10}$.

19.

Ans: Multiply by 8: $8x^2-22x+15=(2x-3)(4x-5)=0$. Roots: $x=\dfrac{3}{2}$ and $x=\dfrac{5}{4}$.

20.

Ans: $4x^2-4a^2x+(a^4-b^4) = [2x-(a^2+b^2)][2x-(a^2-b^2)]=0$. Roots: $x=\dfrac{a^2+b^2}{2}$ and $x=\dfrac{a^2-b^2}{2}$.

21.

Ans: $4\sqrt{3}\,x^2+5x-2\sqrt{3}=4\sqrt{3}\,x^2+8x-3x-2\sqrt{3}=4x(\sqrt{3}\,x+2)-\sqrt{3}(\sqrt{3}\,x+2)=(4x-\sqrt{3})(\sqrt{3}\,x+2)=0$. Roots: $x=\dfrac{\sqrt{3}}{4}$ and $x=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}$.

22.

Ans: LHS $= \dfrac{(x+2)+(2(x+1))}{(x+1)(x+2)} = \dfrac{3x+4}{(x+1)(x+2)}$. Equation: $\dfrac{3x+4}{(x+1)(x+2)} = \dfrac{4}{x+4}$. Cross-multiply: $(3x+4)(x+4)=4(x+1)(x+2)$. $3x^2+16x+16=4x^2+12x+8$. $x^2-4x-8=0$ (wait — let me redo). $4(x^2+3x+2)=3x^2+16x+16 \implies 4x^2+12x+8=3x^2+16x+16 \implies x^2-4x-8=0$. Wait: $(3x+4)(x+4)=3x^2+12x+4x+16=3x^2+16x+16$ and $4(x+1)(x+2)=4(x^2+3x+2)=4x^2+12x+8$. So $3x^2+16x+16=4x^2+12x+8 \implies x^2-4x-8=0$. Hmm, discriminant $=16+32=48$, $x=\dfrac{4\pm4\sqrt{3}}{2}=2\pm2\sqrt{3}$. Roots: $x=2+2\sqrt{3}$ and $x=2-2\sqrt{3}$.

23.

Ans: $x-\dfrac{1}{x}=3 \implies x^2-3x-1=0$. $x=\dfrac{3\pm\sqrt{9+4}}{2}=\dfrac{3\pm\sqrt{13}}{2}$. Roots: $x=\dfrac{3+\sqrt{13}}{2}$ and $x=\dfrac{3-\sqrt{13}}{2}$.

24.

Ans: $\dfrac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}=\dfrac{5}{6}$. Numerator: $(x+1)^2-(x-1)^2=4x$. Denominator: $x^2-1$. So $\dfrac{4x}{x^2-1}=\dfrac{5}{6} \implies 24x=5x^2-5 \implies 5x^2-24x-5=0=(5x+1)(x-5)=0$. Roots: $x=5$ and $x=-\dfrac{1}{5}$.

25.

Ans: Let $t=x^2+3x$. $t^2-2t-8=0=(t-4)(t+2)=0$. $t=4$ or $t=-2$. Case 1: $x^2+3x-4=0 \implies (x+4)(x-1)=0 \implies x=-4$ or $x=1$. Case 2: $x^2+3x+2=0 \implies (x+1)(x+2)=0 \implies x=-1$ or $x=-2$. All roots: $x=1,-1,-2,-4$.

Topic 4 Solutions: Quadratic Formula

26.

Ans: $a=2,b=-7,c=3$. $D=49-24=25$. $x=\dfrac{7\pm5}{4}$. Roots: $x=3$ and $x=\dfrac{1}{2}$.

27.

Ans: $D=1+32=33$. $x=\dfrac{-1\pm\sqrt{33}}{4}$. Roots: $x=\dfrac{-1+\sqrt{33}}{4}$ and $x=\dfrac{-1-\sqrt{33}}{4}$.

28.

Ans: $D=48-48=0$. Equal roots. $x=\dfrac{-4\sqrt{3}}{8}=-\dfrac{\sqrt{3}}{2}$. Both roots: $x=-\dfrac{\sqrt{3}}{2}$.

29.

Ans: $D=45-40=5$. $x=\dfrac{3\sqrt{5}\pm\sqrt{5}}{2}$. Roots: $x=\dfrac{4\sqrt{5}}{2}=2\sqrt{5}$ and $x=\dfrac{2\sqrt{5}}{2}=\sqrt{5}$.

30.

Ans: $D=24-24=0$. Equal roots. $x=\dfrac{2\sqrt{6}}{6}=\dfrac{\sqrt{6}}{3}$. Both roots: $x=\dfrac{\sqrt{6}}{3}$.

31.

Ans: $D=16-20=-4<0$. No real roots exist for $x^2+4x+5=0$.

32.

Ans: Multiply through by $(x-3)(2x+3)$: $2x(2x+3)+(x-3)+3x+9=0 \implies 4x^2+6x+x-3+3x+9=0 \implies 4x^2+10x+6=0 \implies 2x^2+5x+3=0=(2x+3)(x+1)=0$. But $x=-\dfrac{3}{2}$ is excluded. Root: $x=-1$.

Topic 5 Solutions: Nature of Roots – Discriminant

33.

Ans:

$D=9-40=-31<0$. No real roots.
$D=48-48=0$. Real and equal roots.
$D=36-24=12>0$. Two distinct real roots.

34.

Ans: $D=k^2-4k=0 \implies k(k-4)=0$. $k=0$ (rejected, not quadratic) or $\mathbf{k=4}$.

35.

Ans: $D=k^2-24=0 \implies k^2=24 \implies \mathbf{k=\pm2\sqrt{6}}$.

36.

Ans: For $k=1$, the equation becomes $0+0+1=0$, which is false. For $k\neq1$: $D=4(k-1)^2-4(k-1)=4(k-1)[(k-1)-1]=4(k-1)(k-2)=0$. So $k=1$ (rejected) or $\mathbf{k=2}$.

37.

Ans: $kx^2-2kx+6=0$. $D=4k^2-24k=0 \implies 4k(k-6)=0$. $k=0$ (rejected) or $\mathbf{k=6}$.

38.

Ans: For real roots: $D\geq0$. $p^2-36\geq0 \implies p^2\geq36 \implies \mathbf{p\leq-6}$ or $\mathbf{p\geq6}$.

39.

Ans: $(-4)^2+p(-4)-4=0 \implies 16-4p-4=0 \implies 4p=12 \implies \mathbf{p=3}$. Now $x^2+3x+q=0$ has equal roots: $D=9-4q=0 \implies \mathbf{q=\dfrac{9}{4}}$.

40.

Ans: $(k+1)^2-4(k+4)=0 \implies k^2+2k+1-4k-16=0 \implies k^2-2k-15=0 \implies (k-5)(k+3)=0$. $\mathbf{k=5}$ or $\mathbf{k=-3}$. (Check $k=-4$: equation becomes $0$, rejected.)

41.

Ans: Let breadth $= b$, length $= 2b$. $2b\cdot b=800 \implies 2b^2=800 \implies b^2=400 \implies b=20$ m, length $=40$ m. Yes, possible. Length = 40 m, Breadth = 20 m.

Topic 6 Solutions: Word Problems – Numbers & Ages

42.

Ans: Let integers be $n$ and $n+1$. $n(n+1)=306 \implies n^2+n-306=0$. $D=1+1224=1225=35^2$. $n=\dfrac{-1+35}{2}=17$. Integers: 17 and 18.

43.

Ans: Let odd integers be $2n-1$ and $2n+1$. $(2n-1)(2n+1)=483 \implies 4n^2-1=483 \implies n^2=121 \implies n=11$. Integers: 21 and 23.

44.

Ans: Let numbers be $x$ and $27-x$. $x(27-x)=182 \implies x^2-27x+182=0=(x-13)(x-14)=0$. Numbers: 13 and 14.

45.

Ans: Let larger $= x$, smaller $= y$. $x^2-y^2=180$ and $y^2=8x$. Sub: $x^2-8x-180=0=(x-18)(x+10)=0$. $x=18$ (taking positive), $y^2=144 \implies y=12$. Numbers: 18 and 12.

46.

Ans: Let Rohan's age $=x$. Mother's age $=x+26$. $(x+3)(x+26+3)=360 \implies (x+3)(x+29)=360 \implies x^2+32x+87=360 \implies x^2+32x-273=0=(x+39)(x-7)=0$. Rohan's present age = 7 years.

47.

Ans: Let present age $=x$. $\dfrac{1}{x-3}+\dfrac{1}{x+5}=\dfrac{1}{3}$. $\dfrac{2x+2}{x^2+2x-15}=\dfrac{1}{3} \implies 6x+6=x^2+2x-15 \implies x^2-4x-21=0=(x-7)(x+3)=0$. Present age = 7 years.

Topic 7 Solutions: Speed, Distance & Work

48.

Ans: Let speed $= x$ km/h. $\dfrac{360}{x}-\dfrac{360}{x+5}=1 \implies 360(x+5)-360x=x(x+5) \implies 1800=x^2+5x \implies x^2+5x-1800=0=(x+45)(x-40)=0$. Speed = 40 km/h.

49.

Ans: Let smaller tap takes $x$ h, larger takes $x-10$ h. $\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}$ (since $9\frac{3}{8}=\frac{75}{8}$). $\dfrac{2x-10}{x(x-10)}=\dfrac{8}{75} \implies 75(2x-10)=8x(x-10) \implies 150x-750=8x^2-80x \implies 8x^2-230x+750=0 \implies 4x^2-115x+375=0=(4x-15)(x-25)=0$. $x=25$ or $x=3.75$ (rejected since $x-10<0$). Smaller tap: 25 h; Larger tap: 15 h.

50.

Ans: Let stream speed $=x$ km/h. $\dfrac{24}{18-x}-\dfrac{24}{18+x}=1 \implies 24\cdot\dfrac{(18+x)-(18-x)}{(18-x)(18+x)}=1 \implies 48x=324-x^2 \implies x^2+48x-324=0=(x+54)(x-6)=0$. Speed of stream = 6 km/h.

51.

Ans: Let passenger speed $=x$ km/h, express $=x+11$. $\dfrac{132}{x}-\dfrac{132}{x+11}=1 \implies 132\cdot\dfrac{11}{x(x+11)}=1 \implies x^2+11x-1452=0=(x+44)(x-33)=0$. Passenger: 33 km/h; Express: 44 km/h.

52.

Ans: Let original speed $=x$ km/h, time $=\dfrac{600}{x}$ h. New speed $=x-200$, new time $=\dfrac{600}{x-200}$. $\dfrac{600}{x-200}-\dfrac{600}{x}=\dfrac{1}{2} \implies 600\cdot\dfrac{200}{x(x-200)}=\dfrac{1}{2} \implies x^2-200x-240000=0=(x-600)(x+400)=0$. $x=600$ km/h. Duration $=\dfrac{600}{600}=\mathbf{1}$ hour.

Topic 8 Solutions: Geometry & Area

53.

Ans: Let breadth $=b$. Length $=2b+1$. $b(2b+1)=528 \implies 2b^2+b-528=0=(2b+33)(b-16)=0$. $b=16$ m, length $=33$ m. Length = 33 m; Breadth = 16 m.

54.

Ans: Let sides be $x$ and $y$ ($x>y$). $x^2+y^2=468$ and $4x-4y=24 \implies x-y=6 \implies x=y+6$. $(y+6)^2+y^2=468 \implies 2y^2+12y+36=468 \implies y^2+6y-216=0=(y+18)(y-12)=0$. $y=12$ m, $x=18$ m. Sides: 18 m and 12 m.

55.

Ans: Let articles $=n$. Cost each $=2n+3$. $n(2n+3)=90 \implies 2n^2+3n-90=0=(2n+15)(n-6)=0$. Articles produced = 6.

56.

Ans: Let shortest side $=x$. Hypotenuse $=2x+6$. Third side $=(2x+6)-2=2x+4$. Pythagoras: $x^2+(2x+4)^2=(2x+6)^2 \implies x^2+4x^2+16x+16=4x^2+24x+36 \implies x^2-8x-20=0=(x-10)(x+2)=0$. $x=10$ m. Sides: 10 m, 24 m, hypotenuse = 26 m.

57.

Ans: Hypotenuse $=13$ cm. Let side of square $=s$. The altitude from the right angle to hypotenuse $=\dfrac{5\times12}{13}=\dfrac{60}{13}$ cm. Using similar triangles: $s = \dfrac{60}{13} \times \dfrac{13-s}{13} \times 13$... Easier: $s = \dfrac{5\times12}{5+12} = \dfrac{60}{17}$ cm. Side of square $= \dfrac{60}{17}$ cm $\approx 3.53$ cm.

Topic 9 Solutions: Assertion-Reason (A-R)

58.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

$x^2-4=(x-2)(x+2)=0 \implies x=\pm2$ ✓. R provides the general identity $x^2-a^2=(x-a)(x+a)$ which correctly explains why the roots are $\pm2$.

59.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

$D=4-20=-16<0$. No real roots ✓. R correctly states the condition ($D<0$) and explains A.

60.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

$D=(4\sqrt{3})^2-4(3)(4)=48-48=0$ ✓. Roots are equal: $x=\dfrac{4\sqrt{3}}{6}=\dfrac{2\sqrt{3}}{3}$. R correctly explains using $D=0$.

61.

Ans: (b) – Both A and R are true but R is NOT the correct explanation of A.

R is the general condition for equal roots. A states $k=4$ gives equal roots. Check: $D=k^2-4k=16-16=0$ ✓ for $k=4$. Both are true. But R merely gives the formula; it doesn't specifically explain why $k=4$ is correct (it requires applying R, not just stating it). Hence R is true but is not a direct/specific explanation of A — both A and R are true independently.

Topic 10 Solutions: Case Study Questions

62.

Ans:

Let width $=w$. Length $=w+30$. Equation: $w(w+30)=3400 \implies \mathbf{w^2+30w-3400=0}$.
$D=900+13600=14500$. $w=\dfrac{-30+\sqrt{14500}}{2}=\dfrac{-30+50\sqrt{5.8}}{2}$. Factor: $w^2+30w-3400=(w+80)(w-40)$... Check: $80\times(-40)=-3200\neq-3400$. Retry: $(w+85)(w-40)=w^2+45w-3400$. Not right. Using formula: $w=\dfrac{-30+\sqrt{900+13600}}{2}=\dfrac{-30+\sqrt{14500}}{2}$. $\sqrt{14500}\approx120.4$. $w\approx45.2$ m. Exact: $\sqrt{14500}=10\sqrt{145}$. $\mathbf{w=\dfrac{-30+10\sqrt{145}}{2}=5(-3+\sqrt{145})}$ m.
Length $=w+30=5(\sqrt{145}-3)+30=5\sqrt{145}+15$ m.
Perimeter $=2(\text{length}+\text{width})=2(5\sqrt{145}-15+5\sqrt{145}+15)=2\times10\sqrt{145}=20\sqrt{145}\approx241$ m.

63.

Ans:

$x = (2x-60)(x-20) = 2x^2-40x-60x+1200=2x^2-100x+1200$. So $\mathbf{2x^2-101x+1200=0}$.
$D=10201-9600=601$. $x=\dfrac{101\pm\sqrt{601}}{4}$. $\sqrt{601}\approx24.5$. $x\approx\dfrac{101+24.5}{4}\approx31.4$ or $x\approx\dfrac{101-24.5}{4}\approx19.1$. For $x\approx31.4$: Length $\approx2.8$ m, Breadth $\approx11.4$ m.
Cost per m² $=\dfrac{57600}{x}$.
If total cost = ₹40,000: cost/m² $=\dfrac{40000}{x}$. This gives a different area equation. Whether $D\geq0$ determines feasibility.

64.

Ans:

Total chocolates $=7x$. After removing 7: $7x-7$. Distributed among 32: each gets $\dfrac{7x-7}{32}=x$ chocolates. $7x-7=32x \implies -25x=7$. Hmm — re-read: "each student got as many as the number of packets $=7$." So $\dfrac{7x-7}{32}=7 \implies 7x-7=224 \implies 7x=231 \implies \mathbf{x=33}$ chocolates per packet.
$x=33$.
Total chocolates $=7\times33=\mathbf{231}$.
Each student received $\mathbf{7}$ chocolates.

65.

Ans:

$h=0$: $-x^2+25=0 \implies x^2=25 \implies \mathbf{x=\pm5}$. Arch meets road at $x=-5$ m and $x=5$ m (i.e., 5 m on each side of centre).
Maximum height: at $x=0$, $h=-0+25=\mathbf{25}$ m.
Equation: $x^2-25=0$, $a=1,b=0,c=-25$. $D=0+100=\mathbf{100>0}$. Two distinct real roots — the arch crosses road level at two distinct points.
For $h=-5$: $-x^2+25=-5 \implies x^2=30$. $D=4\times30=120>0$. So technically roots exist ($x=\pm\sqrt{30}$), but $h=-5$ is below road level, which is physically impossible for an arch. No, the arch cannot have a height of $-5$ m in context, even though mathematically roots exist.

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