10th Maths Chapter 3 Algebra

10th Maths Chapter 3 Algebra solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter on Algebra moves from linear equations in three variables to GCD and LCM of algebraic expressions, rational expressions, square roots, and quadratic equations solved by factorisation, completing the square, and the formula. It also covers the nature of roots, relations between roots and coefficients, graphs, and matrices. The exercises give students practice in solving, forming equations, and working with matrix operations.

Exercises in this chapter

Key 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

Chapter 3 one-mark practice

  1. A system of three linear equations in three variables is inconsistent if their planes

    1. (1) intersect only at a point
    2. (2) intersect in a line
    3. (3) coincides with each other
    4. (4) do not intersect
    Show answer

    Answer: (4) do not intersect

    1. Inconsistent means the system has no solution.

    2. No solution happens only when the three planes have no common point.

    3. So the planes do not intersect.

    Question 1 of 10
  2. 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 2 of 10
  3. If (x−6)(x - 6) is the HCF of x2−2x−24x^2 - 2x - 24 and x2−kx−6x^2 - kx - 6 then the value of kk is

    1. (1) 33
    2. (2) 55
    3. (3) 66
    4. (4) 88
    Show answer

    Answer: (2) 55

    1. x2−2x−24=(x−6)(x+4)x^2 - 2x - 24 = (x-6)(x+4), so (x−6)(x-6) is a factor.

    2. Since (x−6)(x-6) is the HCF, it must divide x2−kx−6x^2 - kx - 6.

    3. So x=6x = 6 makes it zero: 36−6k−6=036 - 6k - 6 = 0.

    4. 6k=30⇒k=56k = 30 \Rightarrow k = 5.

    Question 3 of 10
  4. 3y−3y÷7y−73y2\dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2} is

    1. (1) 9y7\dfrac{9y}{7}
    2. (2) 9y3(21y−21)\dfrac{9y^3}{(21y - 21)}
    3. (3) 21y2−42y+213y3\dfrac{21y^2 - 42y + 21}{3y^3}
    4. (4) 7(y2−2y+1)y2\dfrac{7(y^2 - 2y + 1)}{y^2}
    Show answer

    Answer: (1) 9y7\dfrac{9y}{7}

    1. 3y−3y÷7y−73y2=3(y−1)y×3y27(y−1)\dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)}

    2. Cancel (y−1)(y-1) and one yy.

    3. =9y7= \dfrac{9y}{7}

    Question 4 of 10
  5. y2+1y2y^2 + \dfrac{1}{y^2} is not equal to

    1. (1) y4+1y2\dfrac{y^4 + 1}{y^2}
    2. (2) (y+1y)2\left(y + \dfrac{1}{y}\right)^2
    3. (3) (y−1y)2+2\left(y - \dfrac{1}{y}\right)^2 + 2
    4. (4) (y+1y)2−2\left(y + \dfrac{1}{y}\right)^2 - 2
    Show answer

    Answer: (2) (y+1y)2\left(y + \dfrac{1}{y}\right)^2

    1. (y+1y)2=y2+2+1y2\left(y + \dfrac{1}{y}\right)^2 = y^2 + 2 + \dfrac{1}{y^2}, which is 22 more than y2+1y2y^2 + \dfrac{1}{y^2}.

    2. Options (A), (C), (D) all simplify to y2+1y2y^2 + \dfrac{1}{y^2}.

    3. So (B) is the one that is not equal.

    Question 5 of 10
  6. xx2−25−8x2+6x+5\dfrac{x}{x^2 - 25} - \dfrac{8}{x^2 + 6x + 5} gives

    1. (1) x2−7x+40(x−5)(x+5)\dfrac{x^2 - 7x + 40}{(x - 5)(x + 5)}
    2. (2) x2+7x+40(x−5)(x+5)(x+1)\dfrac{x^2 + 7x + 40}{(x - 5)(x + 5)(x + 1)}
    3. (3) x2−7x+40(x2−25)(x+1)\dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)}
    4. (4) x2+10(x2−25)(x+1)\dfrac{x^2 + 10}{(x^2 - 25)(x + 1)}
    Show answer

    Answer: (3) x2−7x+40(x2−25)(x+1)\dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)}

    1. x2−25=(x+5)(x−5)x^2 - 25 = (x+5)(x-5) and x2+6x+5=(x+5)(x+1)x^2 + 6x + 5 = (x+5)(x+1).

    2. LCM =(x+5)(x−5)(x+1)= (x+5)(x-5)(x+1).

    3. x(x+1)−8(x−5)(x+5)(x−5)(x+1)=x2+x−8x+40(x2−25)(x+1)\dfrac{x(x+1) - 8(x-5)}{(x+5)(x-5)(x+1)} = \dfrac{x^2 + x - 8x + 40}{(x^2-25)(x+1)}

    4. =x2−7x+40(x2−25)(x+1)= \dfrac{x^2 - 7x + 40}{(x^2-25)(x+1)}

    Question 6 of 10
  7. 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 7 of 10
  8. Which of the following should be added to make x4+64x^4 + 64 a perfect square

    1. (1) 4x24x^2
    2. (2) 16x216x^2
    3. (3) 8x28x^2
    4. (4) −8x2-8x^2
    Show answer

    Answer: (2) 16x216x^2

    1. For (x2+8)2=x4+16x2+64(x^2 + 8)^2 = x^4 + 16x^2 + 64.

    2. Given expression has x4+64x^4 + 64, so the missing middle term is 16x216x^2.

    Question 8 of 10
  9. The solution of (2x−1)2=9\left(2x - 1\right)^2 = 9 is equal to

    1. (1) −1-1
    2. (2) 22
    3. (3) −1, 2-1,\ 2
    4. (4) None of these
    Show answer

    Answer: (3) −1, 2-1,\ 2

    1. (2x−1)2=9⇒2x−1=±3(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3.

    2. 2x−1=3⇒x=22x - 1 = 3 \Rightarrow x = 2.

    3. 2x−1=−3⇒x=−12x - 1 = -3 \Rightarrow x = -1.

    Question 9 of 10
  10. 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 10 of 10

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