Exercise 5.1: Algebraic Formulas and Problems Related to Squares
Determine the square with the help of the formulae (1–16):
Necessary Formulas:
- $(a + b)^2 = a^2 + 2ab + b^2$
- $(a – b)^2 = a^2 – 2ab + b^2$
- $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
1. $a + 5$
Solution: $(a + 5)^2$
$= a^2 + 2 \cdot a \cdot 5 + 5^2$
$= a^2 + 10a + 25$
Answer:$a^2 + 10a + 25$
2. $5x – 7$
Solution: $(5x – 7)^2$
$= (5x)^2 – 2 \cdot 5x \cdot 7 + 7^2$
$= 25x^2 – 70x + 49$
Answer:$25x^2 – 70x + 49$
3. $3a – 11xy$
Solution: $(3a – 11xy)^2$
$= (3a)^2 – 2 \cdot 3a \cdot 11xy + (11xy)^2$
$= 9a^2 – 66axy + 121x^2y^2$
Answer:$9a^2 – 66axy + 121x^2y^2$
4. $5a^2 + 9m^2$
Solution: $(5a^2 + 9m^2)^2$
$= (5a^2)^2 + 2 \cdot 5a^2 \cdot 9m^2 + (9m^2)^2$
$= 25a^4 + 90a^2m^2 + 81m^4$
Answer:$25a^4 + 90a^2m^2 + 81m^4$
5. $55$
Solution: $(55)^2$
$= (50 + 5)^2$
$= (50)^2 + 2 \cdot 50 \cdot 5 + 5^2$
$= 2500 + 500 + 25$
$= 3025$
Answer:$3025$
6. $990$
Solution: $(990)^2$
$= (1000 – 10)^2$
$= (1000)^2 – 2 \cdot 1000 \cdot 10 + 10^2$
$= 1000000 – 20000 + 100$
$= 980100$
Answer:$980100$
7. $xy – 6y$
Solution: $(xy – 6y)^2$
$= (xy)^2 – 2 \cdot xy \cdot 6y + (6y)^2$
$= x^2y^2 – 12xy^2 + 36y^2$
Answer:$x^2y^2 – 12xy^2 + 36y^2$
8. $ax – by$
Solution: $(ax – by)^2$
$= (ax)^2 – 2 \cdot ax \cdot by + (by)^2$
$= a^2x^2 – 2abxy + b^2y^2$
Answer:$a^2x^2 – 2abxy + b^2y^2$
9. $97$
Solution: $(97)^2$
$= (100 – 3)^2$
$= (100)^2 – 2 \cdot 100 \cdot 3 + 3^2$
$= 10000 – 600 + 9$
$= 9409$
Answer:$9409$
10. $2x + y – z$
Solution: $(2x + y – z)^2$
$= \{(2x + y) – z\}^2$
$= (2x + y)^2 – 2 \cdot (2x + y) \cdot z + z^2$
$= \{(2x)^2 + 2 \cdot 2x \cdot y + y^2\} – 2z(2x + y) + z^2$
$= 4x^2 + 4xy + y^2 – 4xz – 2yz + z^2$
$= 4x^2 + y^2 + z^2 + 4xy – 2yz – 4xz$
Answer:$4x^2 + y^2 + z^2 + 4xy – 2yz – 4xz$
11. $2a – b + 3c$
Solution: $(2a – b + 3c)^2$
$= \{(2a – b) + 3c\}^2$
$= (2a – b)^2 + 2 \cdot (2a – b) \cdot 3c + (3c)^2$
$= \{(2a)^2 – 2 \cdot 2a \cdot b + b^2\} + 6c(2a – b) + 9c^2$
$= 4a^2 – 4ab + b^2 + 12ca – 6bc + 9c^2$
$= 4a^2 + b^2 + 9c^2 – 4ab – 6bc + 12ca$
Answer:$4a^2 + b^2 + 9c^2 – 4ab – 6bc + 12ca$
12. $x^2 + y^2 – z^2$
Solution: $(x^2 + y^2 – z^2)^2$
$= \{(x^2 + y^2) – z^2\}^2$
$= (x^2 + y^2)^2 – 2 \cdot (x^2 + y^2) \cdot z^2 + (z^2)^2$
$= \{(x^2)^2 + 2 \cdot x^2 \cdot y^2 + (y^2)^2\} – 2z^2(x^2 + y^2) + z^4$
$= x^4 + 2x^2y^2 + y^4 – 2x^2z^2 – 2y^2z^2 + z^4$
$= x^4 + y^4 + z^4 + 2x^2y^2 – 2y^2z^2 – 2z^2x^2$
Answer:$x^4 + y^4 + z^4 + 2x^2y^2 – 2y^2z^2 – 2z^2x^2$
13. $a – 2b – c$
Solution: $(a – 2b – c)^2$
$= \{(a – 2b) – c\}^2$
$= (a – 2b)^2 – 2 \cdot (a – 2b) \cdot c + c^2$
$= \{a^2 – 2 \cdot a \cdot 2b + (2b)^2\} – 2c(a – 2b) + c^2$
$= a^2 – 4ab + 4b^2 – 2ca + 4bc + c^2$
$= a^2 + 4b^2 + c^2 – 4ab + 4bc – 2ca$
Answer:$a^2 + 4b^2 + c^2 – 4ab + 4bc – 2ca$
14. $3x – 2y + z$
Solution: $(3x – 2y + z)^2$
$= \{(3x – 2y) + z\}^2$
$= (3x – 2y)^2 + 2 \cdot (3x – 2y) \cdot z + z^2$
$= \{(3x)^2 – 2 \cdot 3x \cdot 2y + (2y)^2\} + 2z(3x – 2y) + z^2$
$= 9x^2 – 12xy + 4y^2 + 6zx – 4yz + z^2$
$= 9x^2 + 4y^2 + z^2 – 12xy – 4yz + 6zx$
Answer:$9x^2 + 4y^2 + z^2 – 12xy – 4yz + 6zx$
15. $bc + ca + ab$
Solution: $(bc + ca + ab)^2$
$= \{(bc + ca) + ab\}^2$
$= (bc + ca)^2 + 2 \cdot (bc + ca) \cdot ab + (ab)^2$
$= \{(bc)^2 + 2 \cdot bc \cdot ca + (ca)^2\} + 2ab(bc + ca) + a^2b^2$
$= b^2c^2 + 2abc^2 + c^2a^2 + 2ab^2c + 2a^2bc + a^2b^2$
$= a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2ab^2c + 2abc^2$
Answer:$a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2ab^2c + 2abc^2$
16. $2a^2 + 2b – c^2$
Solution: $(2a^2 + 2b – c^2)^2$
$= \{(2a^2 + 2b) – c^2\}^2$
$= (2a^2 + 2b)^2 – 2 \cdot (2a^2 + 2b) \cdot c^2 + (c^2)^2$
$= \{(2a^2)^2 + 2 \cdot 2a^2 \cdot 2b + (2b)^2\} – 2c^2(2a^2 + 2b) + c^4$
$= 4a^4 + 8a^2b + 4b^2 – 4a^2c^2 – 4bc^2 + c^4$
$= 4a^4 + 4b^2 + c^4 + 8a^2b – 4bc^2 – 4c^2a^2$
Answer:$4a^4 + 4b^2 + c^4 + 8a^2b – 4bc^2 – 4c^2a^2$
Simplify (17–24):
17. $(2a + 1)^2 – 4a(2a + 1) + 4a^2$
Solution: $(2a + 1)^2 – 4a(2a + 1) + 4a^2$
$= (2a + 1)^2 – 2 \cdot (2a + 1) \cdot 2a + (2a)^2$
Let, $2a + 1 = x$ and $2a = y$.
Therefore, Given expression $= x^2 – 2xy + y^2$
$= (x – y)^2$
$= \{(2a + 1) – 2a\}^2$ [Substituting the values of $x$ and $y$]
$= (2a + 1 – 2a)^2$
$= (1)^2$
$= 1$
Answer:$1$
18. $(5a + 3b)^2 + 2(5a + 3b)(4a – 3b) + (4a – 3b)^2$
Solution: $(5a + 3b)^2 + 2(5a + 3b)(4a – 3b) + (4a – 3b)^2$
Let, $5a + 3b = x$ and $4a – 3b = y$.
Therefore, Given expression $= x^2 + 2xy + y^2$
$= (x + y)^2$
$= \{(5a + 3b) + (4a – 3b)\}^2$ [Substituting the values of $x$ and $y$]
$= (5a + 3b + 4a – 3b)^2$
$= (9a)^2$
$= 81a^2$
Answer:$81a^2$
19. $(7a + b)^2 – 2(7a + b)(7a – b) + (7a – b)^2$
Solution: $(7a + b)^2 – 2(7a + b)(7a – b) + (7a – b)^2$
Let, $7a + b = x$ and $7a – b = y$.
Therefore, Given expression $= x^2 – 2xy + y^2$
$= (x – y)^2$
$= \{(7a + b) – (7a – b)\}^2$ [Substituting the values of $x$ and $y$]
$= (7a + b – 7a + b)^2$
$= (2b)^2$
$= 4b^2$
Answer:$4b^2$
20. $(2x + 3y)^2 + 2(2x + 3y)(2x – 3y) + (2x – 3y)^2$
Solution: $(2x + 3y)^2 + 2(2x + 3y)(2x – 3y) + (2x – 3y)^2$
Let, $2x + 3y = a$ and $2x – 3y = b$.
Therefore, Given expression $= a^2 + 2ab + b^2$
$= (a + b)^2$
$= \{(2x + 3y) + (2x – 3y)\}^2$ [Substituting the values of $a$ and $b$]
$= (2x + 3y + 2x – 3y)^2$
$= (4x)^2$
$= 16x^2$
Answer:$16x^2$
21. $(5x – 2)^2 + (5x + 7)^2 – 2(5x – 2)(5x + 7)$
Solution: $(5x – 2)^2 + (5x + 7)^2 – 2(5x – 2)(5x + 7)$
Given expression $= (5x – 2)^2 – 2(5x – 2)(5x + 7) + (5x + 7)^2$
Let, $5x – 2 = a$ and $5x + 7 = b$.
Therefore, Given expression $= a^2 – 2ab + b^2$
$= (a – b)^2$
$= \{(5x – 2) – (5x + 7)\}^2$ [Substituting the values of $a$ and $b$]
$= (5x – 2 – 5x – 7)^2$
$= (-9)^2$
$= 81$
Answer:$81$
22. $(3ab – cd)^2 + 9(cd – ab)^2 + 6(3ab – cd)(cd – ab)$
Solution: $(3ab – cd)^2 + 9(cd – ab)^2 + 6(3ab – cd)(cd – ab)$
$= (3ab – cd)^2 + 6(3ab – cd)(cd – ab) + 9(cd – ab)^2$
$= (3ab – cd)^2 + 2 \cdot (3ab – cd) \cdot 3(cd – ab) + \{3(cd – ab)\}^2$
Let, $3ab – cd = x$ and $3(cd – ab) = y$.
Therefore, Given expression $= x^2 + 2xy + y^2$
$= (x + y)^2$
$= \{(3ab – cd) + 3(cd – ab)\}^2$ [Substituting the values of $x$ and $y$]
$= (3ab – cd + 3cd – 3ab)^2$
$= (2cd)^2$
$= 4c^2d^2$
Answer:$4c^2d^2$
23. $(2x + 5y + 3z)^2 + (5y + 3z – x)^2 – 2(5y + 3z – x)(2x + 5y + 3z)$
Solution: $(2x + 5y + 3z)^2 + (5y + 3z – x)^2 – 2(5y + 3z – x)(2x + 5y + 3z)$
$= (2x + 5y + 3z)^2 – 2(2x + 5y + 3z)(5y + 3z – x) + (5y + 3z – x)^2$
Let, $2x + 5y + 3z = a$ and $5y + 3z – x = b$.
Therefore, Given expression $= a^2 – 2ab + b^2$
$= (a – b)^2$
$= \{(2x + 5y + 3z) – (5y + 3z – x)\}^2$
$= (2x + 5y + 3z – 5y – 3z + x)^2$ [Substituting the values of $a$ and $b$]
$= (3x)^2$
$= 9x^2$
Answer:$9x^2$
24. $(2a – 3b + 4c)^2 + (2a + 3b – 4c)^2 + 2(2a – 3b + 4c)(2a + 3b – 4c)$
Solution:
$(2a – 3b + 4c)^2 + (2a + 3b – 4c)^2 + 2(2a – 3b + 4c)(2a + 3b – 4c)$
$= (2a – 3b + 4c)^2 + 2(2a – 3b + 4c)(2a + 3b – 4c) + (2a + 3b – 4c)^2$
Let, $2a – 3b + 4c = x$ and $2a + 3b – 4c = y$.
Therefore, Given expression $= x^2 + 2xy + y^2$
$= (x + y)^2$
$= \{(2a – 3b + 4c) + (2a + 3b – 4c)\}^2$ [Substituting the values of $x$ and $y$]
$= (2a – 3b + 4c + 2a + 3b – 4c)^2$
$= (4a)^2$
$= 16a^2$
Answer:$16a^2$
Determine the value (25–28):
25. $25x^2 + 36y^2 – 60xy$, when $x = -4, y = -5$
Solution: Given, $x = -4, y = -5$
Given expression $= 25x^2 + 36y^2 – 60xy$
$= 25x^2 – 60xy + 36y^2$
$= (5x)^2 – 2 \cdot 5x \cdot 6y + (6y)^2$
$= (5x – 6y)^2$
$= \{5(-4) – 6(-5)\}^2$ [Substituting the values of $x$ and $y$]
$= (-20 + 30)^2$
$= (10)^2$
$= 100$
Answer:$100$
26. $16a^2 – 24ab + 9b^2$, when $a = 7, b = 6$
Solution: Given, $a = 7, b = 6$
Given expression $= 16a^2 – 24ab + 9b^2$
$= (4a)^2 – 2 \cdot 4a \cdot 3b + (3b)^2$
$= (4a – 3b)^2$
$= \{4(7) – 3(6)\}^2$ [Substituting the values of $a$ and $b$]
$= (28 – 18)^2$
$= (10)^2$
$= 100$
Answer:$100$
27. $9x^2 + 30x + 25$, when $x = -2$
Solution: Given, $x = -2$
Given expression $= 9x^2 + 30x + 25$
$= (3x)^2 + 2 \cdot 3x \cdot 5 + 5^2$
$= (3x + 5)^2$
$= \{3(-2) + 5\}^2$ [Substituting the value of $x$]
$= (-6 + 5)^2$
$= (-1)^2$
$= 1$
Answer:$1$
28. $81a^2 + 18ac + c^2$, when $a = 7, c = -67$
Solution: Given, $a = 7, c = -67$
Given expression $= 81a^2 + 18ac + c^2$
$= (9a)^2 + 2 \cdot 9a \cdot c + c^2$
$= (9a + c)^2$
$= \{9(7) + (-67)\}^2$ [Substituting the values of $a$ and $c$]
$= (63 – 67)^2$
$= (-4)^2$
$= 16$
Answer:$16$
29. If $a – b = 7$ and $ab = 3$, show that $(a + b)^2 = 61$.
Solution: Given, $a – b = 7$ and $ab = 3$
L.H.S. $= (a + b)^2$
$= (a – b)^2 + 4ab$
$= (7)^2 + 4 \cdot 3$
$= 49 + 12$
$= 61$
$= \text{R.H.S.}$
Therefore, $(a + b)^2 = 61$(Shown)
30. If $a + b = 5$ and $ab = 12$, show that $a^2 + b^2 = 1$.
Solution: Given, $a + b = 5$ and $ab = 12$
L.H.S. $= a^2 + b^2$
$= (a + b)^2 – 2ab$
$= (5)^2 – 2 \cdot 12$
$= 25 – 24$
$= 1$
$= \text{R.H.S.}$
Therefore, $a^2 + b^2 = 1$(Shown)
31. If $x + \frac{1}{x} = 5$, prove that $\left(x^2 – \frac{1}{x^2}\right)^2 = 525$.
Solution: Given, $x + \frac{1}{x} = 5$
We know,
$\left(x – \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 – 4 \cdot x \cdot \frac{1}{x}$
or, $\left(x – \frac{1}{x}\right)^2 = 5^2 – 4$
or, $\left(x – \frac{1}{x}\right)^2 = 25 – 4$
or, $\left(x – \frac{1}{x}\right)^2 = 21$
Now, L.H.S. $= \left(x^2 – \frac{1}{x^2}\right)^2$
$= \left\{\left(x + \frac{1}{x}\right)\left(x – \frac{1}{x}\right)\right\}^2$
$= \left(x + \frac{1}{x}\right)^2 \left(x – \frac{1}{x}\right)^2$
$= (5)^2 \cdot 21$
$= 25 \cdot 21$
$= 525$
$= \text{R.H.S.}$
Therefore, $\left(x^2 – \frac{1}{x^2}\right)^2 = 525$(Proved)
32. If $a + b = 8$ and $a – b = 4$, $ab =$ what?
Solution: Given, $a + b = 8$ and $a – b = 4$
We know,
$ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$
$= \left(\frac{8}{2}\right)^2 – \left(\frac{4}{2}\right)^2$
$= (4)^2 – (2)^2$
$= 16 – 4$
$= 12$
Answer:$12$
33. If $x + y = 7$ and $xy = 10$, what is the value of $x^2 + y^2 + 5xy$?
Solution: Given, $x + y = 7$ and $xy = 10$
Given expression $= x^2 + y^2 + 5xy$
$= (x + y)^2 – 2xy + 5xy$
$= (x + y)^2 + 3xy$
$= (7)^2 + 3(10)$ [Substituting values]
$= 49 + 30$
$= 79$
Answer:$79$
34. If $m + \frac{1}{m} = 2$, show that $m^4 + \frac{1}{m^4} = 2$.
Solution: Given, $m + \frac{1}{m} = 2$
L.H.S. $= m^4 + \frac{1}{m^4}$
$= (m^2)^2 + \left(\frac{1}{m^2}\right)^2$
$= \left(m^2 + \frac{1}{m^2}\right)^2 – 2 \cdot m^2 \cdot \frac{1}{m^2}$
$= \left\{\left(m + \frac{1}{m}\right)^2 – 2 \cdot m \cdot \frac{1}{m}\right\}^2 – 2$
$= \{(2)^2 – 2\}^2 – 2$
$= (4 – 2)^2 – 2$
$= (2)^2 – 2$
$= 4 – 2$
$= 2$
$= \text{R.H.S.}$
Therefore, $m^4 + \frac{1}{m^4} = 2$(Shown)