Search

Samacheer Kalvi 10th Maths Exercise 3.11 Solutions

10th Maths Exercise 3.11 solutions (Chapter 3 Algebra) with step-by-step answers for all 7 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance7 answers · full steps below
  1. Q1 (i)x=23, 23x=\dfrac{2}{3},\ \dfrac{2}{3}
  2. Q1 (ii)x=3, −1x=3,\ -1
  3. Q2 (i)x=2, 12x=2,\ \dfrac{1}{2}
  4. Q2 (ii)f=3+32, 3−32f=\dfrac{3+\sqrt{3}}{\sqrt{2}},\ \dfrac{3-\sqrt{3}}{\sqrt{2}}
  5. Q2 (iii)y=233, −1y=\dfrac{23}{3},\ -1
  6. Q2 (iv)y=a+b6, a−b6y=\dfrac{a+b}{6},\ \dfrac{a-b}{6}
  7. Q33.753.75 seconds

Q1. Solve the following quadratic equations by completing the square method

(i)

9x2−12x+4=09x^2 - 12x + 4 = 0

Show steps (7)

  1. 9x2−12x+4=09x^2-12x+4=0

  2. Divide by 99: x2−43x+49=0x^2-\dfrac{4}{3}x+\dfrac{4}{9}=0

  3. x2−43x=−49x^2-\dfrac{4}{3}x=-\dfrac{4}{9}

  4. Half of 43\dfrac{4}{3} is 23\dfrac{2}{3}; add (23)2=49\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9} on both sides.

  5. x2−43x+49=−49+49x^2-\dfrac{4}{3}x+\dfrac{4}{9}=-\dfrac{4}{9}+\dfrac{4}{9}

  6. (x−23)2=0\left(x-\dfrac{2}{3}\right)^2=0

  7. x−23=0x-\dfrac{2}{3}=0

Answer

x=23, 23x=\dfrac{2}{3},\ \dfrac{2}{3}

(ii)

5x+7x−1=3x+2\dfrac{5x + 7}{x - 1} = 3x + 2

Show steps (10)

  1. 5x+7x−1=3x+2\dfrac{5x+7}{x-1}=3x+2

  2. Cross multiply: 5x+7=(3x+2)(x−1)5x+7=(3x+2)(x-1)

  3. 5x+7=3x2−x−25x+7=3x^2-x-2

  4. 3x2−6x−9=03x^2-6x-9=0

  5. Divide by 33: x2−2x−3=0x^2-2x-3=0

  6. x2−2x=3x^2-2x=3

  7. Add 12=11^2=1 on both sides: x2−2x+1=3+1x^2-2x+1=3+1

  8. (x−1)2=4(x-1)^2=4

  9. x−1=±2x-1=\pm 2

  10. x=1+2=3x=1+2=3 or x=1−2=−1x=1-2=-1

Answer

x=3, −1x=3,\ -1

Q2. Solve the following quadratic equations by formula method

(i)

2x2−5x+2=02x^2 - 5x + 2 = 0

  1. 2x2−5x+2=02x^2-5x+2=0, so a=2, b=−5, c=2a=2,\ b=-5,\ c=2

  2. x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

  3. b2−4ac=25−4(2)(2)=25−16=9b^2-4ac=25-4(2)(2)=25-16=9

  4. x=5±94=5±34x=\dfrac{5\pm\sqrt{9}}{4}=\dfrac{5\pm 3}{4}

  5. x=84=2x=\dfrac{8}{4}=2 or x=24=12x=\dfrac{2}{4}=\dfrac{1}{2}

Answer

x=2, 12x=2,\ \dfrac{1}{2}

(ii)

2f2−6f+32=0\sqrt{2}f^2 - 6f + 3\sqrt{2} = 0

  1. 2f2−6f+32=0\sqrt{2}f^2-6f+3\sqrt{2}=0, so a=2, b=−6, c=32a=\sqrt{2},\ b=-6,\ c=3\sqrt{2}

  2. b2−4ac=36−4(2)(32)=36−24=12b^2-4ac=36-4(\sqrt{2})(3\sqrt{2})=36-24=12

  3. f=6±1222=6±2322f=\dfrac{6\pm\sqrt{12}}{2\sqrt{2}}=\dfrac{6\pm 2\sqrt{3}}{2\sqrt{2}}

  4. f=3±32f=\dfrac{3\pm\sqrt{3}}{\sqrt{2}}

  5. Rationalising: f=32±62f=\dfrac{3\sqrt{2}\pm\sqrt{6}}{2}

Answer

f=3+32, 3−32f=\dfrac{3+\sqrt{3}}{\sqrt{2}},\ \dfrac{3-\sqrt{3}}{\sqrt{2}}

(iii)

3y2−20y−23=03y^2 - 20y - 23 = 0

  1. 3y2−20y−23=03y^2-20y-23=0, so a=3, b=−20, c=−23a=3,\ b=-20,\ c=-23

  2. b2−4ac=400−4(3)(−23)=400+276=676b^2-4ac=400-4(3)(-23)=400+276=676

  3. y=20±6766=20±266y=\dfrac{20\pm\sqrt{676}}{6}=\dfrac{20\pm 26}{6}

  4. y=466=233y=\dfrac{46}{6}=\dfrac{23}{3} or y=−66=−1y=\dfrac{-6}{6}=-1

Answer

y=233, −1y=\dfrac{23}{3},\ -1

(iv)

36y2−12ay+(a2−b2)=036y^2 - 12ay + (a^2 - b^2) = 0

  1. 36y2−12ay+(a2−b2)=036y^2-12ay+(a^2-b^2)=0, so A=36, B=−12a, C=a2−b2A=36,\ B=-12a,\ C=a^2-b^2

  2. B2−4AC=144a2−144(a2−b2)=144b2B^2-4AC=144a^2-144(a^2-b^2)=144b^2

  3. y=12a±144b272=12a±12b72y=\dfrac{12a\pm\sqrt{144b^2}}{72}=\dfrac{12a\pm 12b}{72}

  4. y=a±b6y=\dfrac{a\pm b}{6}

Answer

y=a+b6, a−b6y=\dfrac{a+b}{6},\ \dfrac{a-b}{6}

Q3. A ball rolls down a slope and travels a distance d=t2−0.75td = t^2 - 0.75t feet …

A ball rolls down a slope and travels a distance d=t2−0.75td = t^2 - 0.75t feet in tt seconds. Find the time when the distance travelled by the ball is 11.2511.25 feet.

Show steps (8)

  1. Distance d=t2−0.75td=t^2-0.75t and d=11.25d=11.25

  2. t2−0.75t=11.25t^2-0.75t=11.25

  3. Multiply by 44: 4t2−3t−45=04t^2-3t-45=0

  4. a=4, b=−3, c=−45a=4,\ b=-3,\ c=-45

  5. b2−4ac=9+720=729b^2-4ac=9+720=729

  6. t=3±7298=3±278t=\dfrac{3\pm\sqrt{729}}{8}=\dfrac{3\pm 27}{8}

  7. t=308=3.75t=\dfrac{30}{8}=3.75 or t=−248=−3t=\dfrac{-24}{8}=-3

  8. Time cannot be negative, so t=3.75t=3.75 seconds.

Answer

3.753.75 seconds

Quick check

  1. Find the matrix XX if 2X+(1357)=(5795)2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}

    1. (1) (−2−22−1)\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}
    2. (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}
    3. (3) (1222)\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}
    4. (4) (2122)\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}
    Show answer

    Answer: (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    1. 2X=(5795)−(1357)=(444−2)2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}

    2. X=12(444−2)=(222−1)X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    Question 1 of 5

  2. The values of aa and bb if 4x4−24x3+76x2+ax+b4x^4 - 24x^3 + 76x^2 + ax + b is a perfect square are

    1. (1) 100,120100,120
    2. (2) 10,1210,12
    3. (3) −120,100-120,100
    4. (4) 12,1012,10
    Show answer

    Answer: (3) −120,100-120,100

    1. Let 4x4−24x3+76x2+ax+b=(2x2−6x+c)24x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2.

    2. Expanding: 4x4−24x3+(36+4c)x2−12cx+c24x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2.

    3. Compare x2x^2 terms: 36+4c=76⇒c=1036 + 4c = 76 \Rightarrow c = 10.

    4. So a=−12(10)=−120a = -12(10) = -120 and b=102=100b = 10^2 = 100.

    Question 2 of 5

  3. The solution of the system x+y−3z=−6x + y - 3z = -6, −7y+7z=7-7y + 7z = 7, 3z=93z = 9 is

    1. (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3
    2. (2) x=−1, y=2, z=3x = -1,\ y = 2,\ z = 3
    3. (3) x=−1, y=−2, z=3x = -1,\ y = -2,\ z = 3
    4. (4) x=1, y=−2, z=3x = 1,\ y = -2,\ z = 3
    Show answer

    Answer: (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3

    1. From 3z=93z = 9, z=3z = 3.

    2. Put z=3z=3 in −7y+7z=7-7y + 7z = 7: −7y+21=7⇒y=2-7y + 21 = 7 \Rightarrow y = 2.

    3. Put y=2, z=3y=2,\ z=3 in x+y−3z=−6x + y - 3z = -6: x+2−9=−6⇒x=1x + 2 - 9 = -6 \Rightarrow x = 1.

    Question 3 of 5

  4. If AA is a 2×32 \times 3 matrix and BB is a 3×43 \times 4 matrix, how many columns does ABAB have

    1. (1) 33
    2. (2) 44
    3. (3) 22
    4. (4) 55
    Show answer

    Answer: (2) 44

    1. A2×3B3×4A_{2\times 3} B_{3\times 4} gives order 2×42 \times 4.

    2. So ABAB has 44 columns.

    Question 4 of 5

  5. The square root of 256x8y4z1025x6y6z6\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6} is equal to

    1. (1) 165∣x2z4y2∣\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|
    2. (2) 16∣y2x2z4∣16\left|\dfrac{y^2}{x^2 z^4}\right|
    3. (3) 165∣yxz2∣\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|
    4. (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|
    Show answer

    Answer: (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|

    1. 256x8y4z1025x6y6z6=256 x2z425 y2\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}

    2. Taking square root: 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|

    Question 5 of 5
Chapter 3 formulas
  • Roots of a quadratic equation

    ax2+bx+c=0,  (a≠0)  ⇒  x=−b±b2−4ac2aax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • Sum of the roots

    α+β=−ba=−Co-efficient of xCo-efficient of x2\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}

  • Product of the roots

    αβ=ca=Constant termCo-efficient of x2\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}

  • Quadratic equation formed from its roots

    x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

  • Discriminant

    Δ=b2−4ac\Delta = b^2 - 4ac

  • Nature of roots: real and unequal

    Δ>0  ⇒  the roots are real and unequal\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}

  • Nature of roots: real and equal

    Δ=0  ⇒  the roots are real and equal\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}

  • Nature of roots: no real roots

    Δ<0  ⇒  there are no real roots\Delta < 0 \;\Rightarrow\; \text{there are no real roots}

  • Cases for a system of linear equations in three variables

    (i) Unique solution(ii) Infinitely many solutions(iii) No solution\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}

  • LCM of algebraic expressions

    LCM=expression of lowest degree that each given expression exactly divides\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}

  • Quadratic polynomial and its zeroes

    p(x)=ax2+bx+c  (a≠0)  ⇒  p(x) has atmost two zeroes, which lie on the x-axisp(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}

  • Order of a matrix

    Order of A=(Number of rows)×(Number of columns)=m×n\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n

  • Row matrix (row vector)

    A=(a11a12⋯a1n)1×nA = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}

  • Column matrix (column vector)

    A=(a11a21⋮am1)m×1A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}

  • Square matrix

    Number of rows=Number of columns,  order n×n\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n

  • Zero matrix (null matrix)

    aij=0 for all i, ja_{ij} = 0 \text{ for all } i,\, j

  • Transpose of a matrix

    (AT)ij=aji(rows and columns of A interchanged)\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})

  • Diagonal matrix

    aij=0 for i≠j(square matrix)a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})

  • Scalar matrix

    aij={k,i=j0,i≠ja_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}

  • Identity matrix (unit matrix)

    aij={1,i=j0,i≠ja_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}

  • Lower triangular matrix

    aij=0 for i<j(all entries above the leading diagonal are zero)a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})

  • Upper triangular matrix

    aij=0 for i>j(all entries below the leading diagonal are zero)a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})

  • Equality of matrices

    A=B  ⟺  same order and aij=bij for all i, jA = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j

Answers checked against the Samacheer Kalvi textbook · Updated

Spotted a mistake? Email us at [email protected] and we will fix it.