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Samacheer Kalvi 10th Maths Exercise 3.8 Solutions

10th Maths Exercise 3.8 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)x4−12x3+42x2−36x+9=∣x2−6x+3∣\sqrt{x^4-12x^3+42x^2-36x+9}=|x^2-6x+3|
  2. Q1 (ii)4x4−28x3+37x2+42x+9=∣2x2−7x−3∣\sqrt{4x^4-28x^3+37x^2+42x+9}=|2x^2-7x-3|
  3. Q1 (iii)16x4+8x2+1=4x2+1\sqrt{16x^4+8x^2+1}=4x^2+1
  4. Q1 (iv)121x4−198x3−183x2+216x+144=∣11x2−9x−12∣\sqrt{121x^4-198x^3-183x^2+216x+144}=|11x^2-9x-12|
  5. Q2 (i)a=49, b=−42a=49,\ b=-42
  6. Q2 (ii)a=144, b=264a=144,\ b=264
  7. Q3 (i)m=30, n=9m=30,\ n=9
  8. Q3 (ii)m=24, n=−32m=24,\ n=-32

Q1. Find the square root of the following polynomials by division method:

(i)

x4−12x3+42x2−36x+9x^4 - 12x^3 + 42x^2 - 36x + 9

Show steps (7)

  1. Write the polynomial in descending powers: x4−12x3+42x2−36x+9x^4-12x^3+42x^2-36x+9.

  2. x4=x2\sqrt{x^4}=x^2 → first term of the root is x2x^2; remainder −12x3+42x2−36x+9-12x^3+42x^2-36x+9.

  3. Double the quotient: 2x22x^2. Divide −12x3÷2x2=−6x-12x^3\div 2x^2=-6x → second term is −6x-6x.

  4. (2x2−6x)(−6x)=−12x3+36x2(2x^2-6x)(-6x)=-12x^3+36x^2; remainder 6x2−36x+96x^2-36x+9.

  5. Double x2−6xx^2-6x → 2x2−12x2x^2-12x. Divide 6x2÷2x2=36x^2\div 2x^2=3 → third term is 33.

  6. (2x2−12x+3)(3)=6x2−36x+9(2x^2-12x+3)(3)=6x^2-36x+9; remainder 00.

  7. So the square root is x2−6x+3x^2-6x+3.

Answer

x4−12x3+42x2−36x+9=∣x2−6x+3∣\sqrt{x^4-12x^3+42x^2-36x+9}=|x^2-6x+3|

(ii)

37x2−28x3+4x4+42x+937x^2 - 28x^3 + 4x^4 + 42x + 9

Show steps (7)

  1. Arrange in descending powers: 4x4−28x3+37x2+42x+94x^4-28x^3+37x^2+42x+9.

  2. 4x4=2x2\sqrt{4x^4}=2x^2 → first term is 2x22x^2; remainder −28x3+37x2+42x+9-28x^3+37x^2+42x+9.

  3. Double the quotient: 4x24x^2. Divide −28x3÷4x2=−7x-28x^3\div 4x^2=-7x → second term is −7x-7x.

  4. (4x2−7x)(−7x)=−28x3+49x2(4x^2-7x)(-7x)=-28x^3+49x^2; remainder −12x2+42x+9-12x^2+42x+9.

  5. Double 2x2−7x2x^2-7x → 4x2−14x4x^2-14x. Divide −12x2÷4x2=−3-12x^2\div 4x^2=-3 → third term is −3-3.

  6. (4x2−14x−3)(−3)=−12x2+42x+9(4x^2-14x-3)(-3)=-12x^2+42x+9; remainder 00.

  7. So the square root is 2x2−7x−32x^2-7x-3.

Answer

4x4−28x3+37x2+42x+9=∣2x2−7x−3∣\sqrt{4x^4-28x^3+37x^2+42x+9}=|2x^2-7x-3|

(iii)

16x4+8x2+116x^4 + 8x^2 + 1

Show steps (6)

  1. Write with all terms: 16x4+0x3+8x2+0x+116x^4+0x^3+8x^2+0x+1.

  2. 16x4=4x2\sqrt{16x^4}=4x^2 → first term is 4x24x^2; remainder 0x3+8x2+0x+10x^3+8x^2+0x+1.

  3. Double the quotient: 8x28x^2. Divide 0x3÷8x2=00x^3\div 8x^2=0 → second term is 00.

  4. Double 4x2+04x^2+0 → 8x28x^2. Divide 8x2÷8x2=18x^2\div 8x^2=1 → third term is 11.

  5. (8x2+1)(1)=8x2+1(8x^2+1)(1)=8x^2+1; remainder 00.

  6. So the square root is 4x2+14x^2+1.

Answer

16x4+8x2+1=4x2+1\sqrt{16x^4+8x^2+1}=4x^2+1

(iv)

121x4−198x3−183x2+216x+144121x^4 - 198x^3 - 183x^2 + 216x + 144

Show steps (7)

  1. 121x4−198x3−183x2+216x+144121x^4-198x^3-183x^2+216x+144.

  2. 121x4=11x2\sqrt{121x^4}=11x^2 → first term is 11x211x^2; remainder −198x3−183x2+216x+144-198x^3-183x^2+216x+144.

  3. Double the quotient: 22x222x^2. Divide −198x3÷22x2=−9x-198x^3\div 22x^2=-9x → second term is −9x-9x.

  4. (22x2−9x)(−9x)=−198x3+81x2(22x^2-9x)(-9x)=-198x^3+81x^2; remainder −264x2+216x+144-264x^2+216x+144.

  5. Double 11x2−9x11x^2-9x → 22x2−18x22x^2-18x. Divide −264x2÷22x2=−12-264x^2\div 22x^2=-12 → third term is −12-12.

  6. (22x2−18x−12)(−12)=−264x2+216x+144(22x^2-18x-12)(-12)=-264x^2+216x+144; remainder 00.

  7. So the square root is 11x2−9x−1211x^2-9x-12.

Answer

121x4−198x3−183x2+216x+144=∣11x2−9x−12∣\sqrt{121x^4-198x^3-183x^2+216x+144}=|11x^2-9x-12|

Q2. Find the values of aa and bb if the following polynomials are …

Find the values of aa and bb if the following polynomials are perfect squares:

(i)

4x4−12x3+37x2+bx+a4x^4 - 12x^3 + 37x^2 + bx + a

  1. Let the square root be 2x2−3x+c2x^2-3x+c, since 4x4=2x2\sqrt{4x^4}=2x^2 and −12x3÷4x2=−3x-12x^3\div 4x^2=-3x.

  2. Expand: (2x2−3x+c)2=4x4−12x3+(9+4c)x2−6cx+c2(2x^2-3x+c)^2=4x^4-12x^3+(9+4c)x^2-6cx+c^2.

  3. Compare x2x^2 terms: 9+4c=37⇒c=79+4c=37\Rightarrow c=7.

  4. Compare xx terms: b=−6c=−42b=-6c=-42.

  5. Compare constants: a=c2=49a=c^2=49.

Answer

a=49, b=−42a=49,\ b=-42

(ii)

ax4+bx3+361x2+220x+100ax^4 + bx^3 + 361x^2 + 220x + 100

  1. Start from the constant end: let the square root be px2+qx+10px^2+qx+10, since 100=10\sqrt{100}=10.

  2. Expand: (px2+qx+10)2=p2x4+2pq x3+(q2+20p)x2+20q x+100(px^2+qx+10)^2=p^2x^4+2pq\,x^3+(q^2+20p)x^2+20q\,x+100.

  3. Compare xx terms: 20q=220⇒q=1120q=220\Rightarrow q=11.

  4. Compare x2x^2 terms: q2+20p=361⇒121+20p=361⇒p=12q^2+20p=361\Rightarrow 121+20p=361\Rightarrow p=12.

  5. Then a=p2=144a=p^2=144 and b=2pq=2(12)(11)=264b=2pq=2(12)(11)=264.

Answer

a=144, b=264a=144,\ b=264

Q3. Find the values of mm and nn if the following polynomials are …

Find the values of mm and nn if the following polynomials are perfect sqaures:

(i)

36x4−60x3+61x2−mx+n36x^4 - 60x^3 + 61x^2 - mx + n

  1. Let the square root be 6x2−5x+c6x^2-5x+c, since 36x4=6x2\sqrt{36x^4}=6x^2 and −60x3÷12x2=−5x-60x^3\div 12x^2=-5x.

  2. Expand: (6x2−5x+c)2=36x4−60x3+(25+12c)x2−10cx+c2(6x^2-5x+c)^2=36x^4-60x^3+(25+12c)x^2-10cx+c^2.

  3. Compare x2x^2 terms: 25+12c=61⇒c=325+12c=61\Rightarrow c=3.

  4. Compare xx terms: −m=−10c=−30⇒m=30-m=-10c=-30\Rightarrow m=30.

  5. Compare constants: n=c2=9n=c^2=9.

Answer

m=30, n=9m=30,\ n=9

(ii)

x4−8x3+mx2+nx+16x^4 - 8x^3 + mx^2 + nx + 16

  1. Let the square root be x2−4x+cx^2-4x+c, since x4=x2\sqrt{x^4}=x^2 and −8x3÷2x2=−4x-8x^3\div 2x^2=-4x.

  2. Expand: (x2−4x+c)2=x4−8x3+(16+2c)x2−8cx+c2(x^2-4x+c)^2=x^4-8x^3+(16+2c)x^2-8cx+c^2.

  3. Compare constants: c2=16⇒c=4c^2=16\Rightarrow c=4 (taking the positive value).

  4. Compare x2x^2 terms: m=16+2c=16+8=24m=16+2c=16+8=24.

  5. Compare xx terms: n=−8c=−32n=-8c=-32.

Answer

m=24, n=−32m=24,\ n=-32

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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