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Samacheer Kalvi 10th Maths Exercise 3.13 Solutions

10th Maths Exercise 3.13 solutions (Chapter 3 Algebra) with step-by-step answers for all 10 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance7 answers · full steps below
  1. Q1 (i)Δ=1>0\Delta = 1 > 0, so the roots are real, unequal and rational.
  2. Q1 (ii)Δ=5>0\Delta = 5 > 0, so the roots are real, unequal and irrational.
  3. Q1 (iii)Δ=−15<0\Delta = -15 < 0, so the roots are not real (imaginary).
  4. Q1 (iv)Δ=0\Delta = 0, so the roots are real and equal.
  5. Q1 (v)Δ=0\Delta = 0, so the roots are real and equal.
  6. Q2 (i)k=2k = 2 or k=3k = 3
  7. Q2 (ii)k=19k = \dfrac{1}{9} or k=1k = 1

Q1. Determine the nature of the roots for the following quadratic equations:

(i)

15x2+11x+2=015x^2 + 11x + 2 = 0

  1. a=15, b=11, c=2a=15,\ b=11,\ c=2

  2. Δ=b2−4ac=112−4(15)(2)\Delta=b^2-4ac=11^2-4(15)(2)

  3. =121−120=1>0=121-120=1>0

  4. Δ\Delta is positive and a perfect square.

Answer

Δ=1>0\Delta = 1 > 0, so the roots are real, unequal and rational.

(ii)

x2−x−1=0x^2 - x - 1 = 0

  1. a=1, b=−1, c=−1a=1,\ b=-1,\ c=-1

  2. Δ=b2−4ac=(−1)2−4(1)(−1)\Delta=b^2-4ac=(-1)^2-4(1)(-1)

  3. =1+4=5>0=1+4=5>0

  4. 55 is not a perfect square.

Answer

Δ=5>0\Delta = 5 > 0, so the roots are real, unequal and irrational.

(iii)

2t2−3t+32=0\sqrt{2}t^2 - 3t + 3\sqrt{2} = 0

  1. a=2, b=−3, c=32a=\sqrt{2},\ b=-3,\ c=3\sqrt{2}

  2. Δ=b2−4ac=(−3)2−4(2)(32)\Delta=b^2-4ac=(-3)^2-4(\sqrt{2})(3\sqrt{2})

  3. =9−24=−15<0=9-24=-15<0

Answer

Δ=−15<0\Delta = -15 < 0, so the roots are not real (imaginary).

(iv)

9y2−62y+2=09y^2 - 6\sqrt{2}y + 2 = 0

  1. a=9, b=−62, c=2a=9,\ b=-6\sqrt{2},\ c=2

  2. Δ=b2−4ac=(−62)2−4(9)(2)\Delta=b^2-4ac=(-6\sqrt{2})^2-4(9)(2)

  3. =72−72=0=72-72=0

Answer

Δ=0\Delta = 0, so the roots are real and equal.

(v)

9a2b2x2−24abcdx+16c2d2=09a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0, a≠0a \neq 0, b≠0b \neq 0

  1. a=9a2b2, b=−24abcd, c=16c2d2a=9a^2b^2,\ b=-24abcd,\ c=16c^2d^2

  2. Δ=(−24abcd)2−4(9a2b2)(16c2d2)\Delta=(-24abcd)^2-4(9a^2b^2)(16c^2d^2)

  3. =576a2b2c2d2−576a2b2c2d2=0=576a^2b^2c^2d^2-576a^2b^2c^2d^2=0

Answer

Δ=0\Delta = 0, so the roots are real and equal.

Q2. Find the value(s) of 'kk' for which the roots of the following …

Find the value(s) of 'kk' for which the roots of the following equations are real and equal.

(i)

(5k−6)x2+2kx+1=0(5k - 6)x^2 + 2kx + 1 = 0

Show steps (6)

  1. (5k−6)x2+2kx+1=0(5k-6)x^2+2kx+1=0, so a=5k−6, b=2k, c=1a=5k-6,\ b=2k,\ c=1

  2. Roots real and equal ⇒Δ=0\Rightarrow \Delta=0

  3. (2k)2−4(5k−6)(1)=0(2k)^2-4(5k-6)(1)=0

  4. 4k2−20k+24=04k^2-20k+24=0

  5. k2−5k+6=0k^2-5k+6=0

  6. (k−2)(k−3)=0(k-2)(k-3)=0

Answer

k=2k = 2 or k=3k = 3

(ii)

kx2+(6k+2)x+16=0kx^2 + (6k + 2)x + 16 = 0

Show steps (7)

  1. kx2+(6k+2)x+16=0kx^2+(6k+2)x+16=0, so a=k, b=6k+2, c=16a=k,\ b=6k+2,\ c=16

  2. Roots real and equal ⇒Δ=0\Rightarrow \Delta=0

  3. (6k+2)2−4(k)(16)=0(6k+2)^2-4(k)(16)=0

  4. 36k2+24k+4−64k=036k^2+24k+4-64k=0

  5. 36k2−40k+4=036k^2-40k+4=0

  6. 9k2−10k+1=09k^2-10k+1=0

  7. (9k−1)(k−1)=0(9k-1)(k-1)=0

Answer

k=19k = \dfrac{1}{9} or k=1k = 1

Q3. If the roots of (a−b)x2+(b−c)x+(c−a)=0(a - b)x^2 + (b - c)x + (c - a) = 0 are real and equal, then prove that …

If the roots of (a−b)x2+(b−c)x+(c−a)=0(a - b)x^2 + (b - c)x + (c - a) = 0 are real and equal, then prove that bb, aa, cc are in arithmetic progression.

Show steps (9)

  1. Here aa-term =(a−b)=(a-b), bb-term =(b−c)=(b-c), cc-term =(c−a)=(c-a)

  2. Roots real and equal ⇒Δ=0\Rightarrow \Delta=0

  3. (b−c)2−4(a−b)(c−a)=0(b-c)^2-4(a-b)(c-a)=0

  4. b2−2bc+c2−4(ac−a2−bc+ab)=0b^2-2bc+c^2-4(ac-a^2-bc+ab)=0

  5. b2−2bc+c2−4ac+4a2+4bc−4ab=0b^2-2bc+c^2-4ac+4a^2+4bc-4ab=0

  6. 4a2+b2+c2−4ab−4ac+2bc=04a^2+b^2+c^2-4ab-4ac+2bc=0

  7. (2a−b−c)2=0(2a-b-c)^2=0

  8. 2a−b−c=0⇒2a=b+c2a-b-c=0 \Rightarrow 2a=b+c

  9. So a−b=c−aa-b=c-a, the common difference is the same.

Answer

2a=b+c2a = b + c, hence b, a, cb,\ a,\ c are in arithmetic progression.

Q4. If aa , bb are real then show that the roots of …

If aa, bb are real then show that the roots of the equation (a−b)x2−6(a+b)x−9(a−b)=0(a - b)x^2 - 6(a + b)x - 9(a - b) = 0 are real and unequal.

Show steps (7)

  1. (a−b)x2−6(a+b)x−9(a−b)=0(a-b)x^2-6(a+b)x-9(a-b)=0

  2. Here A=a−b, B=−6(a+b), C=−9(a−b)A=a-b,\ B=-6(a+b),\ C=-9(a-b)

  3. Δ=B2−4AC=36(a+b)2+36(a−b)2\Delta=B^2-4AC=36(a+b)^2+36(a-b)^2

  4. =36[(a+b)2+(a−b)2]=36\big[(a+b)^2+(a-b)^2\big]

  5. =36[2a2+2b2]=72(a2+b2)=36\big[2a^2+2b^2\big]=72(a^2+b^2)

  6. Since a,ba,b are real and a≠ba\neq b (else it is not quadratic), a2+b2>0a^2+b^2>0

  7. Therefore Δ=72(a2+b2)>0\Delta=72(a^2+b^2)>0

Answer

Δ=72(a2+b2)>0\Delta = 72(a^2+b^2) > 0, so the roots are real and unequal.

Q5. If the roots of the equation (c2−ab)x2−2(a2−bc)x+b2−ac=0(c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 are real and equal prove …

If the roots of the equation (c2−ab)x2−2(a2−bc)x+b2−ac=0(c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 are real and equal prove that either a=0a = 0 (or) a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc

Show steps (9)

  1. (c2−ab)x2−2(a2−bc)x+(b2−ac)=0(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0

  2. Roots real and equal ⇒Δ=0\Rightarrow \Delta=0

  3. [−2(a2−bc)]2−4(c2−ab)(b2−ac)=0\big[-2(a^2-bc)\big]^2-4(c^2-ab)(b^2-ac)=0

  4. 4(a2−bc)2−4(c2−ab)(b2−ac)=04(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0

  5. (a2−bc)2−(c2−ab)(b2−ac)=0(a^2-bc)^2-(c^2-ab)(b^2-ac)=0

  6. a4−2a2bc+b2c2−(b2c2−ac3−ab3+a2bc)=0a^4-2a^2bc+b^2c^2-\big(b^2c^2-ac^3-ab^3+a^2bc\big)=0

  7. a4−3a2bc+ab3+ac3=0a^4-3a^2bc+ab^3+ac^3=0

  8. a(a3+b3+c3−3abc)=0a\big(a^3+b^3+c^3-3abc\big)=0

  9. So a=0a=0 or a3+b3+c3−3abc=0a^3+b^3+c^3-3abc=0

Answer

Either a=0a = 0 or a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc.

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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