Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.7

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1. Find the square root of the following.
(i) \( \frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}} \)
Answer: First, simplify the fraction inside the square root. We can divide the numbers and subtract the powers of the same variables.
\[ \sqrt{\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}} = \sqrt{4 \frac{x^{4-8} y^{12-4} z^{16-4}}{1}} \]
Now simplify the exponents and the number:
\[ = \sqrt{4 x^{-4} y^{8} z^{12}} \]
Remember that \( x^{-4} = \frac{1}{x^4} \). So, the expression becomes:
\[ = \sqrt{\frac{4 y^{8} z^{12}}{x^{4}}} \]
Next, take the square root of each term:
\[ = \frac{\sqrt{4} \sqrt{y^{8}} \sqrt{z^{12}}}{\sqrt{x^{4}}} \]
\[ = \frac{2 y^{8/2} z^{12/2}}{x^{4/2}} \]
\[ = \frac{2 y^{4} z^{6}}{x^{2}} \]
The final answer must be given with an absolute value sign to ensure it is positive. This helps maintain the principal square root.
\[ = \left| \frac{2 y^{4} z^{6}}{x^{2}} \right| \]
Since \( 2, y^4, z^6, \) and \( x^2 \) are always positive or zero, the absolute value is not strictly needed for the final simplified expression if we assume real numbers. However, for a square root, it's good practice.
\[ = \frac{2 y^{4} z^{6}}{x^{2}} \]In simple words: First, divide the numbers and subtract the small powers from the big powers for each letter. Then, find the square root of what is left by dividing all the powers by 2.

🎯 Exam Tip: Always simplify the expression inside the square root as much as possible before taking the root. Remember to use absolute values for variables in the final answer if their original power was even, to ensure the principal (non-negative) square root.

Question 1. Find the square root of the following.
(ii) \( \frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}} \)
Answer: We need to find the square root of the given fraction. First, let's look at the numerator and denominator separately to see if they can be written as perfect squares.
For the numerator: \( 7x^2 + 2\sqrt{14}x + 2 \)
This looks like \( (a+b)^2 = a^2 + 2ab + b^2 \).
We can write \( 7x^2 = (\sqrt{7}x)^2 \) and \( 2 = (\sqrt{2})^2 \).
The middle term is \( 2(\sqrt{7}x)(\sqrt{2}) = 2\sqrt{14}x \).
So, the numerator is \( (\sqrt{7}x + \sqrt{2})^2 \).
For the denominator: \( x^2 - \frac{1}{2}x + \frac{1}{16} \)
This looks like \( (a-b)^2 = a^2 - 2ab + b^2 \).
We can write \( x^2 = (x)^2 \) and \( \frac{1}{16} = \left(\frac{1}{4}\right)^2 \).
The middle term is \( -2(x)\left(\frac{1}{4}\right) = -\frac{2}{4}x = -\frac{1}{2}x \).
So, the denominator is \( \left(x - \frac{1}{4}\right)^2 \).
Now, substitute these back into the original expression:
\[ \sqrt{\frac{(\sqrt{7}x + \sqrt{2})^2}{\left(x - \frac{1}{4}\right)^2}} \]
Taking the square root of the numerator and the denominator gives:
\[ = \left| \frac{\sqrt{7}x + \sqrt{2}}{x - \frac{1}{4}} \right| \]
To simplify the denominator further, we can multiply the top and bottom by 4 to remove the fraction:
\[ = \left| \frac{4(\sqrt{7}x + \sqrt{2})}{4\left(x - \frac{1}{4}\right)} \right| \]
\[ = \left| \frac{4\sqrt{7}x + 4\sqrt{2}}{4x - 1} \right| \]In simple words: Look at the top and bottom parts of the fraction separately. Try to write each part as something squared. Once both are squared, you can easily take the square root of the whole fraction. Remember to use absolute values for the final answer.

🎯 Exam Tip: Always recognize and factorize perfect square trinomials (like \( a^2 \pm 2ab + b^2 \)) in both the numerator and denominator. This makes taking the square root much simpler and quicker. Also, pay attention to fractional coefficients in the middle term, as they often come from squaring fractions.

Question 1. Find the square root of the following.
(iii) \( \frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}} \)
Answer: To find the square root, we will simplify the expression inside the root first, just like in a normal fraction.
First, combine the terms with the same base in the denominator:
\( (b-c)^4 \times (b-c)^4 = (b-c)^{4+4} = (b-c)^8 \)
So the expression becomes:
\[ \sqrt{\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{8}(a-b)^{12}}} \]
Now, we can cancel out the \( (b-c)^8 \) term from the numerator and denominator:
\[ = \sqrt{\frac{121(a+b)^{8}(x+y)^{8}}{81(a-b)^{12}}} \]
Now, take the square root of each part (the numbers and each variable term) separately. Remember that \( \sqrt{X^n} = X^{n/2} \).
\[ = \frac{\sqrt{121} \sqrt{(a+b)^{8}} \sqrt{(x+y)^{8}}}{\sqrt{81} \sqrt{(a-b)^{12}}} \]
\[ = \frac{11 (a+b)^{8/2} (x+y)^{8/2}}{9 (a-b)^{12/2}} \]
\[ = \frac{11 (a+b)^{4} (x+y)^{4}}{9 (a-b)^{6}} \]
Finally, for square roots of expressions with variables, it's generally good practice to use absolute value signs to ensure the result is positive, unless specified otherwise. This provides the principal square root.
\[ = \left| \frac{11 (a+b)^{4} (x+y)^{4}}{9 (a-b)^{6}} \right| \]
Since all the powers are even, \( (a+b)^4, (x+y)^4, \) and \( (a-b)^6 \) will always be non-negative, so the absolute value sign primarily ensures the expression is well-defined.In simple words: First, simplify the powers of identical terms in the top and bottom of the fraction. Then, find the square root of the number part and divide the power of each letter part by two. This gives you the simplified form.

🎯 Exam Tip: When simplifying fractions with exponents, remember that \( \frac{X^m}{X^n} = X^{m-n} \). Also, take the square root of the numerical coefficients first, then divide each exponent by 2 for the variable terms. The absolute value symbol is important for guaranteeing a non-negative principal square root.

Question 2. Find the square root of the following
(i) \( 4x^2 + 20x + 25 \)
Answer: We need to find the square root of the expression \( 4x^2 + 20x + 25 \).
This expression is a quadratic trinomial. Let's see if it fits the form of a perfect square \( (a+b)^2 = a^2 + 2ab + b^2 \).
The first term \( 4x^2 \) can be written as \( (2x)^2 \). So, \( a = 2x \).
The last term \( 25 \) can be written as \( (5)^2 \). So, \( b = 5 \).
Now, check the middle term \( 2ab \):
\( 2ab = 2(2x)(5) = 20x \).
This matches the middle term of the given expression.
Therefore, \( 4x^2 + 20x + 25 = (2x + 5)^2 \).
Now, we take the square root of this expression:
\[ \sqrt{4x^2 + 20x + 25} = \sqrt{(2x + 5)^2} \]
\[ = |2x + 5| \]
We use the absolute value sign because the square root of a squared term is always non-negative. For example, \( \sqrt{(-3)^2} = \sqrt{9} = 3 \), not -3.In simple words: This math problem asks you to find what number, when multiplied by itself, gives \( 4x^2 + 20x + 25 \). We see that it is a perfect square, so the answer is the absolute value of \( 2x+5 \).

🎯 Exam Tip: Always try to identify if a quadratic expression is a perfect square trinomial of the form \( (ax \pm b)^2 \). If the first term is a perfect square, the last term is a perfect square, and the middle term is twice the product of their square roots, then it's a perfect square. Don't forget the absolute value in the final answer.

Question 2. Find the square root of the following
(ii) \( 9x^2 - 24xy + 30xz - 40yz + 25z^2 + 16y^2 \)
Answer: We need to find the square root of the given expression. This expression has six terms, which suggests it might be the square of a trinomial like \( (a+b+c)^2 \).
The formula for \( (a+b+c)^2 \) is \( a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \).
Let's try to identify \( a, b, \) and \( c \) from the squared terms:
\( 9x^2 = (3x)^2 \Rightarrow a = 3x \)
\( 16y^2 = (4y)^2 \Rightarrow b = \pm 4y \)
\( 25z^2 = (5z)^2 \Rightarrow c = \pm 5z \)
Now let's use the cross-product terms to determine the signs.
\( 2ab = -24xy \). Since \( a=3x \), \( 2(3x)b = -24xy \Rightarrow 6xb = -24xy \Rightarrow b = -4y \). So, \( b \) is \( -4y \).
\( 2ac = +30xz \). Since \( a=3x \), \( 2(3x)c = +30xz \Rightarrow 6xc = +30xz \Rightarrow c = +5z \). So, \( c \) is \( 5z \).
Let's check the remaining term \( 2bc \) with these values:
\( 2bc = 2(-4y)(5z) = -40yz \). This matches the given expression.
So, the expression is \( (3x - 4y + 5z)^2 \).
Therefore, the square root is:
\[ \sqrt{(3x - 4y + 5z)^2} = |3x - 4y + 5z| \]
The absolute value sign ensures that the square root is the principal (non-negative) root.In simple words: This long expression is actually a single thing multiplied by itself. We find the main parts \( 3x \), \( -4y \), and \( 5z \) by looking at the squared terms and the terms that have two different letters. Once we find these parts, the square root is just the absolute value of their sum.

🎯 Exam Tip: For six-term expressions, always suspect the form \( (a+b+c)^2 \). Identify \( a^2, b^2, c^2 \) first, then use the mixed terms \( 2ab, 2bc, 2ca \) to correctly determine the signs of \( a, b, \) and \( c \). The absolute value is crucial for the final square root.

Question 2. Find the square root of the following
(iii) \( 1+\frac{1}{x^{6}}+\frac{2}{x^{3}} \)
Answer: We need to find the square root of the expression \( 1+\frac{1}{x^{6}}+\frac{2}{x^{3}} \).
Let's rearrange the terms to look like a standard quadratic or perfect square form:
\[ 1 + \frac{2}{x^3} + \frac{1}{x^6} \]
This resembles the form \( a^2 + 2ab + b^2 = (a+b)^2 \).
Here, \( a^2 = 1 \implies a = 1 \).
And \( b^2 = \frac{1}{x^6} \implies b = \sqrt{\frac{1}{x^6}} = \frac{1}{x^{6/2}} = \frac{1}{x^3} \).
Now, let's check the middle term \( 2ab \):
\( 2ab = 2(1)\left(\frac{1}{x^3}\right) = \frac{2}{x^3} \).
This matches the middle term in our expression.
So, the expression \( 1+\frac{1}{x^{6}}+\frac{2}{x^{3}} \) can be written as \( \left(1 + \frac{1}{x^3}\right)^2 \).
Now, we take the square root:
\[ \sqrt{1+\frac{1}{x^{6}}+\frac{2}{x^{3}}} = \sqrt{\left(1 + \frac{1}{x^3}\right)^2} \]
\[ = \left|1 + \frac{1}{x^3}\right| \]
The absolute value is included to ensure the principal (non-negative) square root.In simple words: This expression looks like a perfect square. We can see that 1 is \( 1^2 \) and \( \frac{1}{x^6} \) is \( \left(\frac{1}{x^3}\right)^2 \). The middle term matches. So, the square root is simply the absolute value of \( 1 + \frac{1}{x^3} \).

🎯 Exam Tip: When fractions are involved, recognize that \( \frac{1}{x^6} \) is a perfect square of \( \frac{1}{x^3} \). Always try to rearrange terms to fit the \( (a+b)^2 \) or \( (a-b)^2 \) pattern. The absolute value is essential in the final step of finding the square root of a squared term.

Question 2. Find the square root of the following
(iv) \( (4x^2 - 9x + 2)(7x^2 - 13x - 2)(28x^2 - 3x - 1) \)
Answer: To find the square root of this product, we first need to factorize each quadratic expression.
**Factorizing \( 4x^2 - 9x + 2 \):**
We look for two numbers that multiply to \( 4 \times 2 = 8 \) and add up to -9. These numbers are -8 and -1.
\( 4x^2 - 8x - x + 2 \)
\( = 4x(x - 2) - 1(x - 2) \)
\( = (x - 2)(4x - 1) \)
**Factorizing \( 7x^2 - 13x - 2 \):**
We look for two numbers that multiply to \( 7 \times (-2) = -14 \) and add up to -13. These numbers are -14 and 1.
\( 7x^2 - 14x + x - 2 \)
\( = 7x(x - 2) + 1(x - 2) \)
\( = (x - 2)(7x + 1) \)
**Factorizing \( 28x^2 - 3x - 1 \):**
We look for two numbers that multiply to \( 28 \times (-1) = -28 \) and add up to -3. These numbers are -7 and 4.
\( 28x^2 - 7x + 4x - 1 \)
\( = 7x(4x - 1) + 1(4x - 1) \)
\( = (4x - 1)(7x + 1) \)
Now, substitute these factored forms back into the original product:
\[ (x - 2)(4x - 1) \times (x - 2)(7x + 1) \times (4x - 1)(7x + 1) \]
Group the identical factors:
\[ = (x - 2)^2 (4x - 1)^2 (7x + 1)^2 \]
Now, take the square root of this entire product:
\[ \sqrt{(x - 2)^2 (4x - 1)^2 (7x + 1)^2} \]
\[ = |(x - 2)(4x - 1)(7x + 1)| \]
The absolute value is necessary to ensure the principal square root is non-negative.In simple words: First, break down each of the three long math expressions into simpler parts (factors). You'll notice that some of these simpler parts repeat. Once all expressions are factored, group the repeating parts. Then, take the square root of the whole thing, which means just taking one of each grouped pair of factors, and put absolute value bars around the final answer.

🎯 Exam Tip: The key to solving such problems is to factorize each quadratic expression accurately. Look for common factors after factorization, as this indicates a perfect square. Remember that the square root of a product of squared terms is the absolute value of the product of the terms themselves.

Question 2. Find the square root of the following
(v) \( \left(2x^2 + \frac{17}{6}x + 1\right) \left(\frac{3}{2}x^2 + 4x + 2\right) \left(\frac{4}{3}x^2 + \frac{11}{3}x + 2\right) \)
Answer: We need to find the square root of the product of these three quadratic expressions. First, we will factorize each expression.
**Factorizing \( 2x^2 + \frac{17}{6}x + 1 \):**
To make factorization easier, we can clear the fraction by multiplying by 6: \( \frac{1}{6}(12x^2 + 17x + 6) \).
For \( 12x^2 + 17x + 6 \), we look for two numbers that multiply to \( 12 \times 6 = 72 \) and add up to 17. These numbers are 8 and 9.
\( 12x^2 + 8x + 9x + 6 \)
\( = 4x(3x + 2) + 3(3x + 2) \)
\( = (3x + 2)(4x + 3) \)
So, \( 2x^2 + \frac{17}{6}x + 1 = \frac{1}{6}(3x + 2)(4x + 3) \).
**Factorizing \( \frac{3}{2}x^2 + 4x + 2 \):**
Clear the fraction by multiplying by 2: \( \frac{1}{2}(3x^2 + 8x + 4) \).
For \( 3x^2 + 8x + 4 \), we look for two numbers that multiply to \( 3 \times 4 = 12 \) and add up to 8. These numbers are 6 and 2.
\( 3x^2 + 6x + 2x + 4 \)
\( = 3x(x + 2) + 2(x + 2) \)
\( = (x + 2)(3x + 2) \)
So, \( \frac{3}{2}x^2 + 4x + 2 = \frac{1}{2}(x + 2)(3x + 2) \).
**Factorizing \( \frac{4}{3}x^2 + \frac{11}{3}x + 2 \):**
Clear the fraction by multiplying by 3: \( \frac{1}{3}(4x^2 + 11x + 6) \).
For \( 4x^2 + 11x + 6 \), we look for two numbers that multiply to \( 4 \times 6 = 24 \) and add up to 11. These numbers are 8 and 3.
\( 4x^2 + 8x + 3x + 6 \)
\( = 4x(x + 2) + 3(x + 2) \)
\( = (x + 2)(4x + 3) \)
So, \( \frac{4}{3}x^2 + \frac{11}{3}x + 2 = \frac{1}{3}(x + 2)(4x + 3) \).
Now, substitute these factored forms back into the original product:
\[ \left[\frac{1}{6}(3x + 2)(4x + 3)\right] \times \left[\frac{1}{2}(x + 2)(3x + 2)\right] \times \left[\frac{1}{3}(x + 2)(4x + 3)\right] \]
Multiply the fractional coefficients: \( \frac{1}{6} \times \frac{1}{2} \times \frac{1}{3} = \frac{1}{36} \).
Group the identical factors:
\[ = \frac{1}{36} (3x + 2)^2 (4x + 3)^2 (x + 2)^2 \]
Finally, take the square root of the entire product:
\[ \sqrt{\frac{1}{36} (3x + 2)^2 (4x + 3)^2 (x + 2)^2} \]
\[ = \sqrt{\frac{1}{36}} \sqrt{(3x + 2)^2} \sqrt{(4x + 3)^2} \sqrt{(x + 2)^2} \]
\[ = \frac{1}{6} |(3x + 2)(4x + 3)(x + 2)| \]
The absolute value is included to ensure the principal (non-negative) square root.In simple words: This problem involves multiplying three complex expressions and then finding the square root. First, simplify each of the three complex expressions by factoring them. You can multiply each expression by its denominator to remove fractions first, then factor. After factoring all three, multiply them together. You will see that some factors appear twice. Combine these factors into squared terms, then take the square root of the entire result, putting an absolute value sign around the final simplified expression.

🎯 Exam Tip: When dealing with quadratic expressions containing fractions, multiply by the common denominator to convert them into integer coefficient quadratics before factoring. This simplifies the splitting of the middle term. Remember to divide by the same number later. Grouping identical factors after factorization is crucial for identifying perfect squares for the final square root step. Always include the absolute value for the square root of squared terms.