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Samacheer Kalvi 10th Maths Exercise 3.9 Solutions

10th Maths Exercise 3.9 solutions (Chapter 3 Algebra) with step-by-step answers for all 8 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below
  1. Q1 (i)x2+9x+20=0x^2+9x+20=0
  2. Q1 (ii)3x2−5x+12=03x^2-5x+12=0
  3. Q1 (iii)2x2+3x−2=02x^2+3x-2=0
  4. Q1 (iv)x2+(2−a)2 x+(a+5)2=0x^2+(2-a)^2\,x+(a+5)^2=0
  5. Q2 (i)α+β=−3\alpha+\beta=-3, αβ=−28\alpha\beta=-28
  6. Q2 (ii)α+β=−3\alpha+\beta=-3, αβ=0\alpha\beta=0
  7. Q2 (iii)α+β=−13\alpha+\beta=\dfrac{-1}{3}, αβ=−103\alpha\beta=\dfrac{-10}{3}
  8. Q2 (iv)α+β=13\alpha+\beta=\dfrac{1}{3}, αβ=−43\alpha\beta=\dfrac{-4}{3}

Q1. Determine the quadratic equations, whose sum and product of roots are

(i)

−9, 20-9,\ 20

  1. General form: x2−(sum)x+product=0x^2-(\text{sum})x+\text{product}=0.

  2. Sum =−9=-9, product =20=20.

  3. x2−(−9)x+20=0x^2-(-9)x+20=0

Answer

x2+9x+20=0x^2+9x+20=0

(ii)

53, 4\dfrac{5}{3},\ 4

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

  2. x2−53x+4=0x^2-\dfrac{5}{3}x+4=0

  3. Multiply by 33: 3x2−5x+12=03x^2-5x+12=0

Answer

3x2−5x+12=03x^2-5x+12=0

(iii)

−32, −1\dfrac{-3}{2},\ -1

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

  2. x2−(−32)x+(−1)=0x^2-\left(\dfrac{-3}{2}\right)x+(-1)=0

  3. x2+32x−1=0x^2+\dfrac{3}{2}x-1=0

  4. Multiply by 22: 2x2+3x−2=02x^2+3x-2=0

Answer

2x2+3x−2=02x^2+3x-2=0

(iv)

−(2−a)2, (a+5)2-(2-a)^2,\ (a+5)^2

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

  2. Sum =−(2−a)2=-(2-a)^2, product =(a+5)2=(a+5)^2.

  3. x2−[−(2−a)2]x+(a+5)2=0x^2-\left[-(2-a)^2\right]x+(a+5)^2=0

  4. x2+(2−a)2x+(a+5)2=0x^2+(2-a)^2x+(a+5)^2=0

Answer

x2+(2−a)2 x+(a+5)2=0x^2+(2-a)^2\,x+(a+5)^2=0

Q2. Find the sum and product of the roots for each of the …

Find the sum and product of the roots for each of the following quadratic equations

(i)

x2+3x−28=0x^2 + 3x - 28 = 0

  1. x2+3x−28=0x^2+3x-28=0 gives a=1, b=3, c=−28a=1,\ b=3,\ c=-28.

  2. α+β=−ba=−3\alpha+\beta=\dfrac{-b}{a}=-3

  3. αβ=ca=−28\alpha\beta=\dfrac{c}{a}=-28

Answer

α+β=−3\alpha+\beta=-3, αβ=−28\alpha\beta=-28

(ii)

x2+3x=0x^2 + 3x = 0

  1. x2+3x=0x^2+3x=0 gives a=1, b=3, c=0a=1,\ b=3,\ c=0.

  2. α+β=−ba=−3\alpha+\beta=\dfrac{-b}{a}=-3

  3. αβ=ca=0\alpha\beta=\dfrac{c}{a}=0

Answer

α+β=−3\alpha+\beta=-3, αβ=0\alpha\beta=0

(iii)

3+1a=10a23 + \dfrac{1}{a} = \dfrac{10}{a^2}

  1. 3+1a=10a23+\dfrac{1}{a}=\dfrac{10}{a^2}

  2. Multiply by a2a^2: 3a2+a=103a^2+a=10

  3. General form: 3a2+a−10=03a^2+a-10=0

  4. Here coefficients are 3, 1, −103,\ 1,\ -10.

  5. Sum of roots =−13=\dfrac{-1}{3}, product =−103=\dfrac{-10}{3}

Answer

α+β=−13\alpha+\beta=\dfrac{-1}{3}, αβ=−103\alpha\beta=\dfrac{-10}{3}

(iv)

3y2−y−4=03y^2 - y - 4 = 0

  1. 3y2−y−4=03y^2-y-4=0 gives a=3, b=−1, c=−4a=3,\ b=-1,\ c=-4.

  2. α+β=−ba=13\alpha+\beta=\dfrac{-b}{a}=\dfrac{1}{3}

  3. αβ=ca=−43\alpha\beta=\dfrac{c}{a}=\dfrac{-4}{3}

Answer

α+β=13\alpha+\beta=\dfrac{1}{3}, αβ=−43\alpha\beta=\dfrac{-4}{3}

Quick check

  1. Graph of a linear equation is a _______

    1. (1) straight line
    2. (2) circle
    3. (3) parabola
    4. (4) hyperbola
    Show answer

    Answer: (1) straight line

    1. A linear equation in two variables has degree one, and its graph is a straight line.

    Question 1 of 5

  2. 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 2 of 5

  3. Transpose of a column matrix is

    1. (1) unit matrix
    2. (2) diagonal matrix
    3. (3) column matrix
    4. (4) row matrix
    Show answer

    Answer: (4) row matrix

    1. A column matrix has order m×1m \times 1.

    2. Its transpose has order 1×m1 \times m, which is a row matrix.

    Question 3 of 5

  4. 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 4 of 5

  5. 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 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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