Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.8

Question 1. Find the square root of the following polynomials by division method
(i) \( x^4 - 12x^3 + 42x^2 - 36x + 9 \)

Answer: We apply the long division method to find the square root of the polynomial: \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{x^2 - 6x + 3} \\ \cline{2-5} x^2 & \enclose{longdiv}{x^4 - 12x^3 + 42x^2 - 36x + 9} \\ & x^4 \\ \cline{2-2} 2x^2 - 6x & \qquad -12x^3 + 42x^2 \\ & \qquad -12x^3 + 36x^2 \\ \cline{2-4} 2x^2 - 12x + 3 & \qquad \qquad \quad 6x^2 - 36x + 9 \\ & \qquad \qquad \quad 6x^2 - 36x + 9 \\ \cline{2-4} & \qquad \qquad \qquad \quad 0 \\ \end{array} \] The square root of the given polynomial is \( |x^2 - 6x + 3| \). This method systematically finds the polynomial that, when multiplied by itself, gives the original expression.
In simple words: We divide the polynomial step-by-step, just like long division with numbers. The answer is the polynomial at the top, and we put absolute value bars around it because a square root result can be positive or negative.

🎯 Exam Tip: When using the division method for square roots of polynomials, remember to write the terms in descending powers of x, adding zero coefficients for missing terms. Ensure subtraction signs are carefully handled by changing the signs of all terms being subtracted.

Question 1. (ii) \( 31x^2 - 28x^3 + 4x^4 + 42x + 9 \)

Answer: First, we arrange the polynomial terms in descending order of their powers: \( 4x^4 - 28x^3 + 37x^2 + 42x + 9 \) Now, we perform the long division to find the square root: \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{2x^2 - 7x - 3} \\ \cline{2-5} 2x^2 & \enclose{longdiv}{4x^4 - 28x^3 + 37x^2 + 42x + 9} \\ & 4x^4 \\ \cline{2-2} 4x^2 - 7x & \qquad -28x^3 + 37x^2 \\ & \qquad -28x^3 + 49x^2 \\ \cline{2-4} 4x^2 - 14x - 3 & \qquad \qquad \quad -12x^2 + 42x + 9 \\ & \qquad \qquad \quad -12x^2 + 42x + 9 \\ \cline{2-4} & \qquad \qquad \qquad \quad 0 \\ \end{array} \] The square root of the given polynomial is \( |2x^2 - 7x - 3| \). This rearrangement ensures the division proceeds correctly from the highest power of x.
In simple words: First, put the terms of the polynomial in order, starting with the highest power of x. Then, use the long division method to find its square root. The result is written inside absolute value bars.

🎯 Exam Tip: Always arrange polynomial terms in descending order of powers before attempting division, otherwise the algorithm will not work correctly. Be careful with sign changes during subtraction steps.

Question 1. (iii) \( 16x^4 + 8x^2 + 1 \)

Answer: We include terms with zero coefficients for missing powers to ensure correct alignment: \( 16x^4 + 0x^3 + 8x^2 + 0x + 1 \) Now, we find the square root using long division: \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{4x^2 + 1} \\ \cline{2-5} 4x^2 & \enclose{longdiv}{16x^4 + 0x^3 + 8x^2 + 0x + 1} \\ & 16x^4 \\ \cline{2-2} 8x^2 + 1 & \qquad \quad 8x^2 + 0x + 1 \\ & \qquad \quad 8x^2 + 0x + 1 \\ \cline{2-4} & \qquad \qquad \qquad \quad 0 \\ \end{array} \] The square root of the polynomial is \( |4x^2 + 1| \). Using placeholders with zero coefficients helps maintain the polynomial's structure during division.
In simple words: When a polynomial misses some powers of x (like \( x^3 \) or \( x^1 \)), we add them back with a zero in front. Then, we perform the long division to find the square root, and the answer is put inside absolute value bars.

🎯 Exam Tip: For polynomials with missing terms, it's crucial to insert placeholders (e.g., \( +0x^3 \)) to ensure all powers align correctly during the division process, preventing calculation errors.

Question 1. (iv) \( 121x^4 - 198x^3 - 183x^2 + 216x + 144 \)

Answer: We apply the long division method to find the square root: \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{11x^2 - 9x - 12} \\ \cline{2-5} 11x^2 & \enclose{longdiv}{121x^4 - 198x^3 - 183x^2 + 216x + 144} \\ & 121x^4 \\ \cline{2-2} 22x^2 - 9x & \qquad -198x^3 - 183x^2 \\ & \qquad -198x^3 + 81x^2 \\ \cline{2-4} 22x^2 - 18x - 12 & \qquad \qquad \quad -264x^2 + 216x + 144 \\ & \qquad \qquad \quad -264x^2 + 216x + 144 \\ \cline{2-4} & \qquad \qquad \qquad \quad 0 \\ \end{array} \] The square root of the polynomial is \( |11x^2 - 9x - 12| \). Each step ensures we are correctly subtracting terms to simplify the polynomial.
In simple words: We find the square root by dividing the polynomial in steps. The final answer is the polynomial found at the top, written with absolute value bars.

🎯 Exam Tip: Accuracy in arithmetic and careful management of signs are critical for success in polynomial long division. Double-check each subtraction to avoid errors that propagate through the problem.

Question 2. Find the square root of the expression \( \frac{x^{2}}{y^{2}} - \frac { 10x }{ y } + 27 - \frac { 10y }{ x } + \frac{y^{2}}{x^{2}} \)

Answer: We perform long division for the given expression: \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{\frac{x}{y} - 5 + \frac{y}{x}} \\ \cline{2-5} \frac{x}{y} & \enclose{longdiv}{\frac{x^2}{y^2} - \frac{10x}{y} + 27 - \frac{10y}{x} + \frac{y^2}{x^2}} \\ & \frac{x^2}{y^2} \\ \cline{2-2} \frac{2x}{y} - 5 & \qquad -\frac{10x}{y} + 27 \\ & \qquad -\frac{10x}{y} + 25 \\ \cline{2-4} \frac{2x}{y} - 10 + \frac{2y}{x} & \qquad \qquad \quad 2 - \frac{10y}{x} + \frac{y^2}{x^2} \\ & \qquad \qquad \quad 2 - \frac{10y}{x} + \frac{y^2}{x^2} \\ \cline{2-4} & \qquad \qquad \qquad \quad 0 \\ \end{array} \] The square root of the expression is \( |\frac{x}{y} - 5 + \frac{y}{x}| \). This method works for expressions with fractional terms too, by following the same division steps.
In simple words: We find the square root of this expression by using long division, even though it has fractions. The process is similar, and the answer is the expression we get on top, written within absolute value bars.

🎯 Exam Tip: When dealing with fractional polynomial expressions, remember to treat terms like \( \frac{x}{y} \) and \( \frac{y}{x} \) as distinct variables in the division process. Ensure proper common denominators for addition/subtraction if terms cannot be directly combined.

Question 3. Find the values of a and b if the following polynomials are perfect squares.
(i) \( 4x^4 - 12x^3 + 37x^2 + bx + a \)

Answer: We use the long division method to find the square root. Since the polynomial is a perfect square, the remainder must be zero. \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{2x^2 - 3x + 7} \\ \cline{2-5} 2x^2 & \enclose{longdiv}{4x^4 - 12x^3 + 37x^2 + bx + a} \\ & 4x^4 \\ \cline{2-2} 4x^2 - 3x & \qquad -12x^3 + 37x^2 \\ & \qquad -12x^3 + 9x^2 \\ \cline{2-4} 4x^2 - 6x + 7 & \qquad \qquad \quad 28x^2 + bx + a \\ & \qquad \qquad \quad 28x^2 - 42x + 49 \\ \cline{2-4} & \qquad \qquad \qquad \quad (b+42)x + (a-49) \\ \end{array} \] For the polynomial to be a perfect square, the remainder must be zero.
\( \implies (b+42)x + (a-49) = 0 \)
This means that the coefficients of x and the constant term must both be zero.
\( \implies b+42 = 0 \implies b = -42 \)
\( \implies a-49 = 0 \implies a = 49 \)
Thus, the values are \( a = 49 \) and \( b = -42 \). Finding these coefficients makes the polynomial perfectly divisible.
In simple words: We divide the polynomial to find its square root. Because it's a "perfect square," the leftover part (remainder) must be zero. By setting the remainder to zero, we can find the unknown numbers 'a' and 'b'.

🎯 Exam Tip: For a polynomial to be a perfect square, its long division remainder must identically be zero. Equating the coefficients of the remainder polynomial to zero is the key to solving for unknown constants like 'a' and 'b'.

Question 3. (ii) \( ax^4 + bx^3 + 361x^2 + 220x + 100 \)

Answer: First, we rearrange the polynomial in ascending powers of x, as shown in the solution: \( 100 + 220x + 361x^2 + bx^3 + ax^4 \) Now, we perform the long division. Since it's a perfect square, the remainder will be zero. \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{10 + 11x + 12x^2} \\ \cline{2-5} 10 & \enclose{longdiv}{100 + 220x + 361x^2 + bx^3 + ax^4} \\ & 100 \\ \cline{2-2} 20 + 11x & \qquad 220x + 361x^2 \\ & \qquad 220x + 121x^2 \\ \cline{2-4} 20 + 22x + 12x^2 & \qquad \qquad \quad 240x^2 + bx^3 + ax^4 \\ & \qquad \qquad \quad 240x^2 + 264x^3 + 144x^4 \\ \cline{2-4} & \qquad \qquad \qquad \quad (b-264)x^3 + (a-144)x^4 \\ \end{array} \] For the polynomial to be a perfect square, the remainder must be zero.
\( \implies (b-264)x^3 + (a-144)x^4 = 0 \)
This means that the coefficients of \( x^3 \) and \( x^4 \) must both be zero.
\( \implies b-264 = 0 \implies b = 264 \)
\( \implies a-144 = 0 \implies a = 144 \)
Hence, the values are \( a = 144 \) and \( b = 264 \). This systematic approach helps solve for unknown coefficients.
In simple words: We first put the polynomial terms in order (from smallest power of x to biggest). Then, we do the long division. Since it's a "perfect square," the remainder has to be zero. By making the remainder zero, we find the values of 'a' and 'b'.

🎯 Exam Tip: Polynomial division can be performed in either ascending or descending order of powers. Consistency in ordering and meticulous calculation are essential, especially when solving for unknown coefficients from the remainder.

Question 4. Find the values of m and n if the following expressions are perfect squares.
(i) \( \frac{1}{x^{4}} - \frac{6}{x^{3}} + \frac{13}{x^{2}} + \frac { m }{ x } + n \)

Answer: We perform the long division. Since the expression is a perfect square, the remainder must be zero. \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{\frac{1}{x^2} - \frac{3}{x} + 2} \\ \cline{2-5} \frac{1}{x^2} & \enclose{longdiv}{\frac{1}{x^4} - \frac{6}{x^3} + \frac{13}{x^2} + \frac{m}{x} + n} \\ & \frac{1}{x^4} \\ \cline{2-2} \frac{2}{x^2} - \frac{3}{x} & \qquad -\frac{6}{x^3} + \frac{13}{x^2} \\ & \qquad -\frac{6}{x^3} + \frac{9}{x^2} \\ \cline{2-4} \frac{2}{x^2} - \frac{6}{x} + 2 & \qquad \qquad \quad \frac{4}{x^2} + \frac{m}{x} + n \\ & \qquad \qquad \quad \frac{4}{x^2} - \frac{12}{x} + 4 \\ \cline{2-4} & \qquad \qquad \qquad \quad (\frac{m}{x} + \frac{12}{x}) + (n-4) \\ & \qquad \qquad \qquad \quad \frac{m+12}{x} + (n-4) \\ \end{array} \] For the polynomial to be a perfect square, the remainder must be zero.
\( \implies \frac{m+12}{x} + (n-4) = 0 \)
This implies that the coefficient of \( \frac{1}{x} \) and the constant term must both be zero.
\( \implies m+12 = 0 \implies m = -12 \)
\( \implies n-4 = 0 \implies n = 4 \)
Therefore, the values are \( m = -12 \) and \( n = 4 \). This helps complete the expression into a perfect square.
In simple words: We do long division with the expression. Since it's a "perfect square," the leftover part (remainder) must be zero. By making the parts of the remainder equal to zero, we can find the unknown numbers 'm' and 'n'.

🎯 Exam Tip: When terms are fractions involving x, ensure consistent algebraic operations. Pay close attention to combining like terms, especially when they have different powers of x in the denominator, to correctly determine the remainder's coefficients.

Question 4. (ii) \( x^4 - 8x^3 + mx^2 + nx + 16 \)

Answer: We apply the long division method. For the polynomial to be a perfect square, the remainder must be zero. \[ \require{enclose} \begin{array}{r c c c} \multicolumn{2}{r}{x^2 - 4x + 4} \\ \cline{2-5} x^2 & \enclose{longdiv}{x^4 - 8x^3 + mx^2 + nx + 16} \\ & x^4 \\ \cline{2-2} 2x^2 - 4x & \qquad -8x^3 + mx^2 \\ & \qquad -8x^3 + 16x^2 \\ \cline{2-4} 2x^2 - 8x + 4 & \qquad \qquad \quad (m-16)x^2 + nx + 16 \\ & \qquad \qquad \quad 8x^2 - 32x + 16 \\ \cline{2-4} & \qquad \qquad \qquad \quad (m-16-8)x^2 + (n+32)x + (16-16) \\ & \qquad \qquad \qquad \quad (m-24)x^2 + (n+32)x + 0 \\ \end{array} \] For the polynomial to be a perfect square, the remainder must be zero.
\( \implies (m-24)x^2 + (n+32)x = 0 \)
This implies that the coefficients of \( x^2 \) and x must both be zero.
\( \implies m-24 = 0 \implies m = 24 \)
\( \implies n+32 = 0 \implies n = -32 \)
Thus, the values are \( m = 24 \) and \( n = -32 \). This helps ensure the polynomial meets the perfect square condition.
In simple words: We perform long division. Since the polynomial is a "perfect square," the remainder has to be zero. By setting the parts of the remainder equal to zero, we can find the unknown numbers 'm' and 'n'.

🎯 Exam Tip: When the dividend has unknown coefficients, ensure you correctly subtract the product of the quotient and divisor at each step. The final remainder's coefficients are then used to form equations to solve for the unknowns.