Maharashtra Board Class 7 Maths Chapter 14 Set 50 Algebraic Formulae Expansion of Squares Solutions

Question 1. Expand:
(i) \((5a + 6b)^2\)
(ii) \((\frac{a}{2} + \frac{b}{3})^2\)
(iii) \((2p – 3q)^2\)
(iv) \((x - \frac{2}{x})^2\)
(v) \((ax + by)^2\)
(vi) \((7m – 4)^2\)
(vii) \((x + \frac{1}{2})^2\)
(viii) \((a - \frac{1}{a})^2\)
Answer:
Solution:
(i) \((5a + 6b)^2\)
Here, \(A = 5a\) and \(B = 6b\)
\((5a + 6b)^2 = (5a)^2 + 2 \times 5a \times 6b + (6b)^2\)
\[(A + B)^2 = A^2 + 2AB + B^2\]
\(\therefore (5a + 6b)^2 = 25a^2 + 60ab + 36b^2\)
(ii) \((\frac{a}{2} + \frac{b}{3})^2\)
Here \(A = \frac{a}{2}\) and \(B = \frac{b}{3}\)
\((\frac{a}{2} + \frac{b}{3})^2 = (\frac{a}{2})^2 + 2 \times \frac{a}{2} \times \frac{b}{3} + (\frac{b}{3})^2\)
\[(A + B)^2 = A^2 + 2AB + B^2\]
\(\therefore (\frac{a}{2} + \frac{b}{3})^2 = \frac{a^2}{4} + \frac{ab}{3} + \frac{b^2}{9}\)
(iii) \((2p – 3q)^2\)
Here, \(a = 2p\) and \(b = 3q\)
\((2p – 3q)^2 = (2p)^2 – 2 \times (2p) \times (3q) + (3q)^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(\therefore (2p – 3q)^2 = 4p^2 – 12pq + 9q^2\)
(iv) \((x - \frac{2}{x})^2\)
Here \(a = x\) and \(b = \frac{2}{x}\)
\((x - \frac{2}{x})^2 = x^2 – 2 \times x \times \frac{2}{x} + (\frac{2}{x})^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(\therefore (x - \frac{2}{x})^2 = x^2 – 4 + \frac{4}{x^2}\)
(v) \((ax + by)^2\)
Here, \(A = ax\) and \(B = by\)
\((ax + by)^2 = (ax)^2 + 2 \times ax \times by + (by)^2\)
\[(A + B)^2 = A^2 + 2AB + B^2\]
\(\therefore (ax + by)^2 = a^2x^2 + 2abxy + b^2y^2\)
(vi) \((7m – 4)^2\)
Here, \(a = 7m\) and \(b = 4\)
\((7m – 4)^2 = (7m)^2 – 2 \times 7m \times 4 + 4^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(\therefore (7m – 4)^2 = 49m^2 – 56m + 16\)
(vii) \((x + \frac{1}{2})^2\)
Here \(a = x\) and \(b = \frac{1}{2}\)
\((x + \frac{1}{2})^2 = x^2 + 2 \times x \times \frac{1}{2} + (\frac{1}{2})^2\)
\[(a + b)^2 = a^2 + 2ab + b^2\]
\(\therefore (x + \frac{1}{2})^2 = x^2 + x + \frac{1}{4}\)
(viii) \((a - \frac{1}{a})^2\)
Here \(A = a\) and \(B = \frac{1}{a}\)
\((a - \frac{1}{a})^2 = a^2 – 2 \times a \times \frac{1}{a} + (\frac{1}{a})^2\)
\[(A - B)^2 = A^2 – 2AB + B^2\]
\(\therefore (a - \frac{1}{a})^2 = a^2 – 2 + \frac{1}{a^2}\)
In simple words: This question requires applying the binomial expansion formulas \((a+b)^2 = a^2+2ab+b^2\) and \((a-b)^2 = a^2-2ab+b^2\) to various algebraic expressions. Each sub-part involves identifying 'a' and 'b' and then substituting them into the appropriate formula to expand the square.

🎯 Exam Tip: Remember to correctly identify the 'a' and 'b' terms in each expression, especially with fractions or negative signs, and apply the correct expansion formula. Careful calculation of each term, including the middle term \(2ab\), is crucial for accuracy.

Question 2. Which of the options given below is the square of the binomial
(A) \(64 - \frac{1}{x^2}\)
(B) \(64 + \frac{16}{x} + \frac{1}{x^2}\)
(C) \(64 - \frac{16}{x} + \frac{1}{x^2}\)
(D) \(64 + \frac{16}{x} + \frac{1}{x^2}\)
Answer: (C) \(64 - \frac{16}{x} + \frac{1}{x^2}\)
Solution:
Hint:
\((8 - \frac{1}{x})^2 = 8^2 – 2 \times 8 \times \frac{1}{x} + (\frac{1}{x})^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(= 64 - \frac{16}{x} + \frac{1}{x^2}\)
In simple words: The problem asks to identify which binomial, when squared, results in the given expanded form. By recognizing the pattern of the expression \(64 - \frac{16}{x} + \frac{1}{x^2}\) as a perfect square expansion of the form \((a-b)^2\), we can deduce that it comes from squaring \((8 - \frac{1}{x})\).

🎯 Exam Tip: For multiple-choice questions involving expansions, try to work backward from the expanded form to identify the original binomial. Look for perfect square terms at the ends and check if the middle term matches \(2ab\) (or \(-2ab\)).

Question 3. Of which of the binomials given below is the \(m^2n^2 + 14mnpq + 49p^2q^2\) the expansion?
(A) \((m + n) (p + q)\)
(B) \((mn – pq)\)
(C) \((7mn + pq)\)
(D) \((mn + 7pq)\)
Answer: (D) (mn + 7pq)
Solution:
Hint:
Here, square root of the first term \( = \sqrt{m^2n^2} = mn\)
Square root of the last term \( = \sqrt{49p^2q^2} = 7pq\)
\(\therefore\) Required binomial \( = (mn + 7pq)^2\)
In simple words: This question asks to find the binomial whose square results in the given trinomial. By taking the square root of the first and last terms of the trinomial (\(m^2n^2\) and \(49p^2q^2\)), we identify the components of the binomial as \(mn\) and \(7pq\). The positive middle term \(14mnpq\) indicates an addition, so the binomial is \((mn + 7pq)\).

🎯 Exam Tip: When given a trinomial and asked to find the binomial from which it was expanded, check if the first and last terms are perfect squares. If they are, take their square roots. Then, verify if twice the product of these square roots equals the middle term. The sign of the middle term will determine if it's \((a+b)^2\) or \((a-b)^2\).

Question 4. Use an expansion formula to find the values of:
(i) \((997)^2\)
(ii) \((102)^2\)
(iii) \((97)^2\)
(iv) \((1005)^2\)
Answer:
Solution:
(i) \((997)^2 = (1000 – 3)^2\)
Here, \(a = 1000\) and \(b = 3\)
\((1000 – 3)^2 = (1000)^2 – 2 \times 1000 \times 3 + 3^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(= 1000000 – 6000 + 9\)
\(= 994009\)
\(\therefore (997)^2 = 994009\)
(ii) \((102)^2 = (100 + 2)^2\)
Here, \(a = 100\) and \(b = 2\)
\((100 + 2)^2 = (100)^2 + 2 \times 100 \times 2 + 2^2\)
\[(a + b)^2 = a^2 + 2ab + b^2\]
\(= 10000 + 400 + 4\)
\(= 10404\)
\(\therefore (102)^2 = 10404\)
(iii) \((97)^2 = (100 – 3)^2\)
Here, \(a = 100\) and \(b = 3\)
\((100 – 3)^2 = (100)^2 – 2 \times 100 \times 3 + 3^2\)
\[(a - b)^2 = a^2 – 2ab + b^2\]
\(= 10000 - 600 + 9\)
\(= 9409\)
\(\therefore (97)^2 = 9409\)
(iv) \((1005)^2 = (1000 + 5)^2\)
Here, \(a = 1000\) and \(b = 5\)
\((1000 + 5)^2 = (1000)^2 + 2 \times 1000 \times 5 + 5^2\)
\[(a + b)^2 = a^2 + 2ab + b^2\]
\(= 1000000 + 10000 + 25\)
\(= 1010025\)
\(\therefore (1005)^2 = 1010025\)
In simple words: This question demonstrates how to simplify squaring numbers close to multiples of 10, 100, or 1000 by using the binomial expansion formulas \((a+b)^2\) or \((a-b)^2\). By expressing the numbers as a sum or difference of a round number and a small digit, calculations become much easier and can often be done mentally or with fewer steps than direct multiplication.

🎯 Exam Tip: When calculating squares of numbers ending in 7, 8, 9, 1, 2, 3, or 4, try to express them as \((X-Y)^2\) or \((X+Y)^2\) where X is a multiple of 10 or 100. This method simplifies computation, reduces calculation errors, and is a key technique for quick mental math and competitive exams.

Maharashtra Board Class 7 Maths Chapter 14 Algebraic Formulae – Expansion Of Squares Practice Set 50 Intext Questions And Activities

Question 1. Use the given values to verify the formulae for squares of binomials. (Textbook pg. no. 92)
(i) \(a = -7, b = 8\)
(ii) \(a = 11, b = 3\)
(iii) \(a = 2.5, b = 1.2\)
Answer:
Solution:
(i) \(a = -7, b = 8\)
\((a + b)^2 = (-7 + 8)^2 = (1)^2 = 1\)
\(a^2 + 2ab + b^2 = (-7)^2 + 2 \times (-7) \times 8 + 8^2 = 49 - 112 + 64 = 1\)
\(\therefore (a + b)^2 = a^2 + 2ab + b^2\)
\((a – b)^2 = (-7 – 8)^2 = (-15)^2 = 225\)
\(a^2 – 2ab + b^2 = (-7)^2 – 2 \times (-7) \times 8 + (8)^2 = 49 + 112 + 64 = 225\)
\(\therefore (a – b)^2 = a^2 – 2ab + b^2\)
(ii) \(a = 11, b = 3\)
\((a + b)^2 = (11 + 3)^2 = (14)^2 = 196\)
\(a^2 + 2ab + b^2 = 11^2 + 2 \times 11 \times 3 + 3^2 = 121 + 66 + 9 = 196\)
\(\therefore (a + b)^2 = a^2 + 2ab + b^2\)
\((a – b)^2 = (11 – 3)^2 = 8^2 = 64\)
\(a^2 – 2ab + b^2 = 11^2 – 2 \times 11 \times 3 + 3^2 = 121 - 66 + 9 = 64\)
\(\therefore (a – b)^2 = a^2 – 2ab + b^2\)
(iii) \(a = 2.5, b = 1.2\)
\((a + b)^2 = (2.5 + 1.2)^2 = (3.7)^2 = 13.69\)
\(a^2 + 2ab + b^2 = (2.5)^2 + 2 \times 2.5 \times 1.2 + (1.2)^2 = 6.25 + 6 + 1.44 = 13.69\)
\(\therefore (a + b)^2 = a^2 + 2ab + b^2\)
\((a – b)^2 = (2.5 – 1.2)^2 = (1.3)^2 = 1.69\)
\(a^2 – 2ab + b^2 = (2.5)^2 – 2 \times 2.5 \times 1.2 + (1.2)^2 = 6.25 - 6 + 1.44 = 1.69\)
\(\therefore (a – b)^2 = a^2 – 2ab + b^2\)
In simple words: This question verifies the algebraic identities for the square of a sum \((a+b)^2 = a^2+2ab+b^2\) and the square of a difference \((a-b)^2 = a^2-2ab+b^2\) by substituting specific numerical values for 'a' and 'b'. For each set of values, both sides of the equations are calculated separately to show that they yield identical results, thus proving the formulas.

🎯 Exam Tip: When verifying identities with given values, ensure you substitute the values correctly into both sides of the equation and perform calculations meticulously. This exercise strengthens understanding of how algebraic formulas hold true for different types of numbers, including negative and decimal values.