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Samacheer Kalvi 10th Maths Exercise 3.14 Solutions

10th Maths Exercise 3.14 solutions (Chapter 3 Algebra) with step-by-step answers for all 13 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance13 answers · full steps below
  1. Q1 (i)(α+β)2−2αβ3αβ\dfrac{(\alpha+\beta)^2-2\alpha\beta}{3\alpha\beta}
  2. Q1 (ii)α+β(αβ)2\dfrac{\alpha+\beta}{(\alpha\beta)^2}
  3. Q1 (iii)9αβ−3(α+β)+19\alpha\beta-3(\alpha+\beta)+1
  4. Q1 (iv)(α+β)2−2αβ+3(α+β)αβ\dfrac{(\alpha+\beta)^2-2\alpha\beta+3(\alpha+\beta)}{\alpha\beta}
  5. Q2 (i)75\dfrac{7}{5}
  6. Q2 (ii)2910\dfrac{29}{10}
  7. Q2 (iii)136\dfrac{13}{6}
  8. Q3 (i)x2−44x+16=0x^2-44x+16=0
  9. Q3 (ii)x2−3x−1=0x^2-3x-1=0
  10. Q3 (iii)x2−24x−64=0x^2-24x-64=0
  11. Q4a=15a=15 or a=−15a=-15
  12. Q5a=24a=24 or a=−24a=-24
  13. Q6k=−36k=-36

Q1. Write each of the following expression in terms of α+β\alpha + \beta and αβ\alpha\beta …

Write each of the following expression in terms of α+β\alpha + \beta and αβ\alpha\beta.

(i)

α3β+β3α\dfrac{\alpha}{3\beta} + \dfrac{\beta}{3\alpha}

  1. Take LCM 3αβ3\alpha\beta:

  2. α3β+β3α=α2+β23αβ\frac{\alpha}{3\beta}+\frac{\beta}{3\alpha}=\frac{\alpha^2+\beta^2}{3\alpha\beta}

  3. Use α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta:

  4. =(α+β)2−2αβ3αβ=\frac{(\alpha+\beta)^2-2\alpha\beta}{3\alpha\beta}

Answer

(α+β)2−2αβ3αβ\dfrac{(\alpha+\beta)^2-2\alpha\beta}{3\alpha\beta}

(ii)

1α2β+1β2α\dfrac{1}{\alpha^2\beta} + \dfrac{1}{\beta^2\alpha}

  1. LCM of α2β\alpha^2\beta and β2α\beta^2\alpha is α2β2\alpha^2\beta^2:

  2. 1α2β+1β2α=β+αα2β2\frac{1}{\alpha^2\beta}+\frac{1}{\beta^2\alpha}=\frac{\beta+\alpha}{\alpha^2\beta^2}

  3. =α+β(αβ)2=\frac{\alpha+\beta}{(\alpha\beta)^2}

Answer

α+β(αβ)2\dfrac{\alpha+\beta}{(\alpha\beta)^2}

(iii)

(3α−1)(3β−1)(3\alpha - 1)(3\beta - 1)

  1. Expand:

  2. (3α−1)(3β−1)=9αβ−3α−3β+1(3\alpha-1)(3\beta-1)=9\alpha\beta-3\alpha-3\beta+1

  3. =9αβ−3(α+β)+1=9\alpha\beta-3(\alpha+\beta)+1

Answer

9αβ−3(α+β)+19\alpha\beta-3(\alpha+\beta)+1

(iv)

α+3β+β+3α\dfrac{\alpha + 3}{\beta} + \dfrac{\beta + 3}{\alpha}

  1. Take LCM αβ\alpha\beta:

  2. α+3β+β+3α=α(α+3)+β(β+3)αβ\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha}=\frac{\alpha(\alpha+3)+\beta(\beta+3)}{\alpha\beta}

  3. =α2+β2+3(α+β)αβ=\frac{\alpha^2+\beta^2+3(\alpha+\beta)}{\alpha\beta}

  4. Put α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta:

  5. =(α+β)2−2αβ+3(α+β)αβ=\frac{(\alpha+\beta)^2-2\alpha\beta+3(\alpha+\beta)}{\alpha\beta}

Answer

(α+β)2−2αβ+3(α+β)αβ\dfrac{(\alpha+\beta)^2-2\alpha\beta+3(\alpha+\beta)}{\alpha\beta}

Q2. The roots of the equation 2x2−7x+5=02x^2 - 7x + 5 = 0 are α\alpha and β\beta . Without …

The roots of the equation 2x2−7x+5=02x^2 - 7x + 5 = 0 are α\alpha and β\beta. Without solving for the roots, find

(i)

1α+1β\dfrac{1}{\alpha} + \dfrac{1}{\beta}

  1. For 2x2−7x+5=02x^2-7x+5=0:

  2. α+β=72,αβ=52\alpha+\beta=\frac{7}{2},\qquad \alpha\beta=\frac{5}{2}

  3. 1α+1β=α+βαβ=7/25/2=75\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7/2}{5/2}=\frac{7}{5}

Answer

75\dfrac{7}{5}

(ii)

αβ+βα\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}

  1. α+β=72, αβ=52\alpha+\beta=\dfrac{7}{2},\ \alpha\beta=\dfrac{5}{2}

  2. αβ+βα=α2+β2αβ=(α+β)2−2αβαβ\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}

  3. =(72)2−2(52)52=494−552=\frac{\left(\frac{7}{2}\right)^2-2\left(\frac{5}{2}\right)}{\frac{5}{2}}=\frac{\frac{49}{4}-5}{\frac{5}{2}}

  4. =29/45/2=2910=\frac{29/4}{5/2}=\frac{29}{10}

Answer

2910\dfrac{29}{10}

(iii)

α+2β+2+β+2α+2\dfrac{\alpha + 2}{\beta + 2} + \dfrac{\beta + 2}{\alpha + 2}

Show steps (6)

  1. α+β=72, αβ=52\alpha+\beta=\dfrac{7}{2},\ \alpha\beta=\dfrac{5}{2}

  2. α+2β+2+β+2α+2=(α+2)2+(β+2)2(α+2)(β+2)\frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2}=\frac{(\alpha+2)^2+(\beta+2)^2}{(\alpha+2)(\beta+2)}

  3. Numerator =α2+β2+4(α+β)+8=\alpha^2+\beta^2+4(\alpha+\beta)+8

  4. =(494−5)+4(72)+8=294+22=1174=\left(\frac{49}{4}-5\right)+4\left(\frac{7}{2}\right)+8=\frac{29}{4}+22=\frac{117}{4}

  5. Denominator =αβ+2(α+β)+4=52+7+4=272=\alpha\beta+2(\alpha+\beta)+4=\dfrac{5}{2}+7+4=\dfrac{27}{2}

  6. ∴ 117/427/2=11754=136\therefore\ \frac{117/4}{27/2}=\frac{117}{54}=\frac{13}{6}

Answer

136\dfrac{13}{6}

Q3. The roots of the equation x2+6x−4=0x^2 + 6x - 4 = 0 are α\alpha , β\beta . Find …

The roots of the equation x2+6x−4=0x^2 + 6x - 4 = 0 are α\alpha, β\beta. Find the quadratic equation whose roots are

(i)

α2\alpha^2 and β2\beta^2

Show steps (7)

  1. For x2+6x−4=0x^2+6x-4=0:

  2. α+β=−6,αβ=−4\alpha+\beta=-6,\qquad \alpha\beta=-4

  3. Sum of new roots =α2+β2=(α+β)2−2αβ=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta

  4. =36−2(−4)=44=36-2(-4)=44

  5. Product =α2β2=(αβ)2=16=\alpha^2\beta^2=(\alpha\beta)^2=16

  6. Equation: x2−(sum)x+(product)=0x^2-(\text{sum})x+(\text{product})=0

  7. x2−44x+16=0x^2-44x+16=0

Answer

x2−44x+16=0x^2-44x+16=0

(ii)

2α\dfrac{2}{\alpha} and 2β\dfrac{2}{\beta}

  1. α+β=−6, αβ=−4\alpha+\beta=-6,\ \alpha\beta=-4

  2. Sum =2α+2β=2(α+β)αβ=2(−6)−4=3=\dfrac{2}{\alpha}+\dfrac{2}{\beta}=\dfrac{2(\alpha+\beta)}{\alpha\beta}=\dfrac{2(-6)}{-4}=3

  3. Product =4αβ=4−4=−1=\dfrac{4}{\alpha\beta}=\dfrac{4}{-4}=-1

  4. x2−3x−1=0x^2-3x-1=0

Answer

x2−3x−1=0x^2-3x-1=0

(iii)

α2β\alpha^2\beta and β2α\beta^2\alpha

  1. α+β=−6, αβ=−4\alpha+\beta=-6,\ \alpha\beta=-4

  2. Sum =α2β+β2α=αβ(α+β)=(−4)(−6)=24=\alpha^2\beta+\beta^2\alpha=\alpha\beta(\alpha+\beta)=(-4)(-6)=24

  3. Product =α2β⋅β2α=(αβ)3=(−4)3=−64=\alpha^2\beta\cdot\beta^2\alpha=(\alpha\beta)^3=(-4)^3=-64

  4. x2−24x−64=0x^2-24x-64=0

Answer

x2−24x−64=0x^2-24x-64=0

Q4. If α\alpha , β\beta are the roots of 7x2+ax+2=07x^2 + ax + 2 = 0 and if β−α=−137\beta - \alpha = \dfrac{-13}{7} …

If α\alpha, β\beta are the roots of 7x2+ax+2=07x^2 + ax + 2 = 0 and if β−α=−137\beta - \alpha = \dfrac{-13}{7}. Find the values of aa.

Show steps (8)

  1. For 7x2+ax+2=07x^2+ax+2=0:

  2. α+β=−a7,αβ=27\alpha+\beta=\frac{-a}{7},\qquad \alpha\beta=\frac{2}{7}

  3. Given β−α=−137\beta-\alpha=\dfrac{-13}{7}

  4. (β−α)2=(α+β)2−4αβ(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta

  5. 16949=a249−87\frac{169}{49}=\frac{a^2}{49}-\frac{8}{7}

  6. 16949=a2−5649\frac{169}{49}=\frac{a^2-56}{49}

  7. a2−56=169 ⇒ a2=225a^2-56=169\ \Rightarrow\ a^2=225

  8. a=±15a=\pm 15

Answer

a=15a=15 or a=−15a=-15

Q5. If one root of the equation 2y2−ay+64=02y^2 - ay + 64 = 0 is twice the other then …

If one root of the equation 2y2−ay+64=02y^2 - ay + 64 = 0 is twice the other then find the values of aa.

  1. Let the roots of 2y2−ay+64=02y^2-ay+64=0 be α\alpha and 2α2\alpha.

  2. Product: α(2α)=642=32\alpha(2\alpha)=\dfrac{64}{2}=32

  3. 2α2=32 ⇒ α2=16 ⇒ α=±42\alpha^2=32\ \Rightarrow\ \alpha^2=16\ \Rightarrow\ \alpha=\pm4

  4. Sum: α+2α=a2\alpha+2\alpha=\dfrac{a}{2}, so 3α=a23\alpha=\dfrac{a}{2} and a=6αa=6\alpha

  5. If α=4\alpha=4, a=24a=24; if α=−4\alpha=-4, a=−24a=-24

Answer

a=24a=24 or a=−24a=-24

Q6. If one root of the equation 3x2+kx+81=03x^2 + kx + 81 = 0 (having real roots) is the …

If one root of the equation 3x2+kx+81=03x^2 + kx + 81 = 0 (having real roots) is the square of the other then find kk.

Show steps (6)

  1. Let the roots of 3x2+kx+81=03x^2+kx+81=0 be α\alpha and α2\alpha^2.

  2. Product: α⋅α2=813=27\alpha\cdot\alpha^2=\dfrac{81}{3}=27

  3. α3=27 ⇒ α=3\alpha^3=27\ \Rightarrow\ \alpha=3

  4. Sum: α+α2=−k3\alpha+\alpha^2=\dfrac{-k}{3}

  5. 3+9=−k3 ⇒ 12=−k33+9=\frac{-k}{3}\ \Rightarrow\ 12=\frac{-k}{3}

  6. k=−36k=-36

Answer

k=−36k=-36

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

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