Exercise 4.1: Algebraic Formulas and Square Problems
1. Find the square of the following expressions with the help of formulae:
Necessary Formulae:
- $(a + b)^2 = a^2 + 2ab + b^2$
- $(a – b)^2 = a^2 – 2ab + b^2$
- $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
(a) $5a + 7b$
Solution: $(5a + 7b)^2$
$= (5a)^2 + 2 \cdot 5a \cdot 7b + (7b)^2$
$= 25a^2 + 70ab + 49b^2$(Ans.)
(b) $6x + 3$
Solution: $(6x + 3)^2$
$= (6x)^2 + 2 \cdot 6x \cdot 3 + (3)^2$
$= 36x^2 + 36x + 9$(Ans.)
(c) $7p – 2q$
Solution: $(7p – 2q)^2$
$= (7p)^2 – 2 \cdot 7p \cdot 2q + (2q)^2$
$= 49p^2 – 28pq + 4q^2$(Ans.)
(d) $ax – by$
Solution: $(ax – by)^2$
$= (ax)^2 – 2 \cdot ax \cdot by + (by)^2$
$= a^2x^2 – 2abxy + b^2y^2$(Ans.)
(e) $x^3 + xy$
Solution: $(x^3 + xy)^2$
$= (x^3)^2 + 2 \cdot x^3 \cdot xy + (xy)^2$
$= x^6 + 2x^4y + x^2y^2$(Ans.)
(f) $11a – 12b$
Solution: $(11a – 12b)^2$
$= (11a)^2 – 2 \cdot 11a \cdot 12b + (12b)^2$
$= 121a^2 – 264ab + 144b^2$(Ans.)
(g) $6x^2y – 5xy^2$
Solution: $(6x^2y – 5xy^2)^2$
$= (6x^2y)^2 – 2 \cdot 6x^2y \cdot 5xy^2 + (5xy^2)^2$
$= 36x^4y^2 – 60x^3y^3 + 25x^2y^4$(Ans.)
(h) $- x – y$
Solution: $(- x – y)^2$
$= \{-(x + y)\}^2$
$= (x + y)^2$
$= x^2 + 2xy + y^2$(Ans.)
(i) $- xyz – abc$
Solution: $(- xyz – abc)^2$
$= \{-(xyz + abc)\}^2$
$= (xyz + abc)^2$
$= (xyz)^2 + 2 \cdot xyz \cdot abc + (abc)^2$
$= x^2y^2z^2 + 2abcxyz + a^2b^2c^2$(Ans.)
(j) $a^2x^3 – b^2y^4$
Solution: $(a^2x^3 – b^2y^4)^2$
$= (a^2x^3)^2 – 2 \cdot a^2x^3 \cdot b^2y^4 + (b^2y^4)^2$
$= a^4x^6 – 2a^2b^2x^3y^4 + b^4y^8$(Ans.)
(k) $108$
Solution: $(108)^2$
$= (100 + 8)^2$
$= (100)^2 + 2 \cdot 100 \cdot 8 + (8)^2$
$= 10000 + 1600 + 64$
$= 11664$(Ans.)
(l) $606$
Solution: $(606)^2$
$= (600 + 6)^2$
$= (600)^2 + 2 \cdot 600 \cdot 6 + (6)^2$
$= 360000 + 7200 + 36$
$= 367236$(Ans.)
(m) $597$
Solution: $(597)^2$
$= (600 – 3)^2$
$= (600)^2 – 2 \cdot 600 \cdot 3 + (3)^2$
$= 360000 – 3600 + 9$
$= 356409$(Ans.)
(n) $a – b + c$
Solution: $(a – b + c)^2$
$= \{a + (- b) + c\}^2$
$= a^2 + (- b)^2 + c^2 + 2 \cdot a \cdot (- b) + 2 \cdot (- b) \cdot c + 2 \cdot c \cdot a$
$= a^2 + b^2 + c^2 – 2ab – 2bc + 2ca$(Ans.)
(o) $ax + b + 2$
Solution: $(ax + b + 2)^2$
$= (ax)^2 + b^2 + (2)^2 + 2 \cdot ax \cdot b + 2 \cdot b \cdot 2 + 2 \cdot 2 \cdot ax$
$= a^2x^2 + b^2 + 4 + 2abx + 4b + 4ax$
$= a^2x^2 + b^2 + 2abx + 4ax + 4b + 4$(Ans.)
(p) $xy + yz – zx$
Solution: $(xy + yz – zx)^2$
$= \{xy + yz + (- zx)\}^2$
$= (xy)^2 + (yz)^2 + (- zx)^2 + 2 \cdot xy \cdot yz + 2 \cdot yz \cdot (- zx) + 2 \cdot (- zx) \cdot xy$
$= x^2y^2 + y^2z^2 + z^2x^2 + 2xy^2z – 2xyz^2 – 2x^2yz$(Ans.)
(q) $3p + 2q – 5r$
Solution: $(3p + 2q – 5r)^2$
$= \{3p + 2q + (- 5r)\}^2$
$= (3p)^2 + (2q)^2 + (- 5r)^2 + 2 \cdot 3p \cdot 2q + 2 \cdot 2q \cdot (- 5r) + 2 \cdot (- 5r) \cdot 3p$
$= 9p^2 + 4q^2 + 25r^2 + 12pq – 20qr – 30rp$(Ans.)
(r) $x^2 – y^2 – z^2$
Solution: $(x^2 – y^2 – z^2)^2$
$= \{x^2 + (- y^2) + (- z^2)\}^2$
$= (x^2)^2 + (- y^2)^2 + (- z^2)^2 + 2 \cdot x^2 \cdot (- y^2) + 2 \cdot (- y^2) \cdot (- z^2) + 2 \cdot (- z^2) \cdot x^2$
$= x^4 + y^4 + z^4 – 2x^2y^2 + 2y^2z^2 – 2z^2x^2$(Ans.)
(s) $7a^2 + 8b^2 – 5c^2$
Solution: $(7a^2 + 8b^2 – 5c^2)^2$
$= \{7a^2 + 8b^2 + (- 5c^2)\}^2$
$= (7a^2)^2 + (8b^2)^2 + (- 5c^2)^2 + 2 \cdot 7a^2 \cdot 8b^2 + 2 \cdot 8b^2 \cdot (- 5c^2) + 2 \cdot (- 5c^2) \cdot 7a^2$
$= 49a^4 + 64b^4 + 25c^4 + 112a^2b^2 – 80b^2c^2 – 70c^2a^2$(Ans.)
2. Simplify:
(a) $(x + y)^2 + 2(x + y)(x – y) + (x – y)^2$
Solution: Let $x + y = a$ and $x – y = b$.
Given expression $= a^2 + 2ab + b^2$
$= (a + b)^2$
$= \{(x + y) + (x – y)\}^2$ [Substituting the values of $a$ and $b$]
$= (x + y + x – y)^2$
$= (2x)^2$
$= 4x^2$(Ans.)
(b) $(2a + 3b)^2 – 2(2a + 3b)(3b – a) + (3b – a)^2$
Solution: Let $2a + 3b = x$ and $3b – a = y$.
Given expression $= x^2 – 2xy + y^2$
$= (x – y)^2$
$= \{(2a + 3b) – (3b – a)\}^2$ [Substituting the values of $x$ and $y$]
$= (2a + 3b – 3b + a)^2$
$= (3a)^2$
$= 9a^2$(Ans.)
(c) $(3x^2 + 7y^2)^2 + 2(3x^2 + 7y^2)(3x^2 – 7y^2) + (3x^2 – 7y^2)^2$
Solution: Let $3x^2 + 7y^2 = a$ and $3x^2 – 7y^2 = b$.
Given expression $= a^2 + 2ab + b^2$
$= (a + b)^2$
$= \{(3x^2 + 7y^2) + (3x^2 – 7y^2)\}^2$ [Substituting the values of $a$ and $b$]
$= (3x^2 + 7y^2 + 3x^2 – 7y^2)^2$
$= (6x^2)^2$
$= 36x^4$(Ans.)
(d) $(8x + y)^2 – (16x + 2y)(5x + y) + (5x + y)^2$
Solution: $(8x + y)^2 – (16x + 2y)(5x + y) + (5x + y)^2$
$= (8x + y)^2 – 2(8x + y)(5x + y) + (5x + y)^2$
Let $8x + y = a$ and $5x + y = b$.
Given expression $= a^2 – 2ab + b^2$
$= (a – b)^2$
$= \{(8x + y) – (5x + y)\}^2$ [Substituting the values of $a$ and $b$]
$= (8x + y – 5x – y)^2$
$= (3x)^2$
$= 9x^2$(Ans.)
(e) $(5x^2 – 3x – 2)^2 + (2 + 5x^2 – 3x)^2 – 2(5x^2 – 3x – 2)(2 + 5x^2 – 3x)$
Solution: $(5x^2 – 3x – 2)^2 + (2 + 5x^2 – 3x)^2 – 2(5x^2 – 3x – 2)(2 + 5x^2 – 3x)$
$= (5x^2 – 3x – 2)^2 – 2(5x^2 – 3x – 2)(5x^2 – 3x + 2) + (5x^2 – 3x + 2)^2$
Let $5x^2 – 3x – 2 = a$ and $5x^2 – 3x + 2 = b$.
Given expression $= a^2 – 2ab + b^2$
$= (a – b)^2$
$= \{(5x^2 – 3x – 2) – (5x^2 – 3x + 2)\}^2$ [Substituting the values of $a$ and $b$]
$= (5x^2 – 3x – 2 – 5x^2 + 3x – 2)^2$
$= (-4)^2$
$= 16$(Ans.)
3. Find the product applying formulae:
Necessary Formulae:
- $(a + b)(a – b) = a^2 – b^2$
- $(x + a)(x + b) = x^2 + (a + b)x + ab$
(a) $(x + 7)(x – 7)$
Solution: $(x + 7)(x – 7)$
$= (x)^2 – (7)^2$
$= x^2 – 49$(Ans.)
(b) $(5x + 13)(5x – 13)$
Solution: $(5x + 13)(5x – 13)$
$= (5x)^2 – (13)^2$
$= 25x^2 – 169$(Ans.)
(c) $(xy + yz)(xy – yz)$
Solution: $(xy + yz)(xy – yz)$
$= (xy)^2 – (yz)^2$
$= x^2y^2 – y^2z^2$(Ans.)
(d) $(ax + b)(ax – b)$
Solution: $(ax + b)(ax – b)$
$= (ax)^2 – (b)^2$
$= a^2x^2 – b^2$(Ans.)
(e) $(a + 3)(a + 4)$
Solution: $(a + 3)(a + 4)$
$= a^2 + (3 + 4)a + (3 \cdot 4)$
$= a^2 + 7a + 12$(Ans.)
(f) $(ax + 3)(ax + 4)$
Solution: $(ax + 3)(ax + 4)$
$= (ax)^2 + (3 + 4)ax + (3 \cdot 4)$
$= a^2x^2 + 7ax + 12$(Ans.)
(g) $(6x + 17)(6x – 13)$
Solution: $(6x + 17)(6x – 13)$
$= \{6x + 17\}\{6x + (-13)\}$
$= (6x)^2 + \{17 + (-13)\} \cdot 6x + \{17 \cdot (-13)\}$
$= 36x^2 + (17 – 13) \cdot 6x – 221$
$= 36x^2 + 4 \cdot 6x – 221$
$= 36x^2 + 24x – 221$(Ans.)
(h) $(a^2 + b^2)(a^2 – b^2)(a^4 + b^4)$
Solution: $(a^2 + b^2)(a^2 – b^2)(a^4 + b^4)$
$= \{(a^2)^2 – (b^2)^2\}(a^4 + b^4)$
$= (a^4 – b^4)(a^4 + b^4)$
$= (a^4)^2 – (b^4)^2$
$= a^8 – b^8$(Ans.)
(i) $(ax – by + cz)(ax + by – cz)$
Solution: $(ax – by + cz)(ax + by – cz)$
$= \{ax – (by – cz)\}\{ax + (by – cz)\}$
$= (ax)^2 – (by – cz)^2$
$= a^2x^2 – \{(by)^2 – 2 \cdot by \cdot cz + (cz)^2\}$
$= a^2x^2 – (b^2y^2 – 2bycz + c^2z^2)$
$= a^2x^2 – b^2y^2 + 2bycz – c^2z^2$(Ans.)
(j) $(3a – 10)(3a – 5)$
Solution: $(3a – 10)(3a – 5)$
$= \{3a + (-10)\}\{3a + (-5)\}$
$= (3a)^2 + \{(-10) + (-5)\} \cdot 3a + \{(-10) \cdot (-5)\}$
$= 9a^2 + (-15) \cdot 3a + 50$
$= 9a^2 – 45a + 50$(Ans.)
(k) $(5a + 2b – 3c)(5a + 2b + 3c)$
Solution: $(5a + 2b – 3c)(5a + 2b + 3c)$
$= \{(5a + 2b) – 3c\}\{(5a + 2b) + 3c\}$
$= (5a + 2b)^2 – (3c)^2$
$= (5a)^2 + 2 \cdot 5a \cdot 2b + (2b)^2 – 9c^2$
$= 25a^2 + 20ab + 4b^2 – 9c^2$(Ans.)
(l) $(ax + by + 5)(ax + by + 3)$
Solution: $(ax + by + 5)(ax + by + 3)$
$= \{(ax + by) + 5\}\{(ax + by) + 3\}$
$= (ax + by)^2 + (5 + 3)(ax + by) + (5 \cdot 3)$
$= \{(ax)^2 + 2 \cdot ax \cdot by + (by)^2\} + 8(ax + by) + 15$
$= a^2x^2 + 2abxy + b^2y^2 + 8ax + 8by + 15$(Ans.)
4. If $a = 4, b = 6$, and $c = 3$, find the value of $4a^2b^2 – 16ab^2c + 16b^2c^2$.
Solution:Given, $a = 4, b = 6$, and $c = 3$.
Given expression $= 4a^2b^2 – 16ab^2c + 16b^2c^2$
$= (2ab)^2 – 2 \cdot (2ab) \cdot (4bc) + (4bc)^2$
$= (2ab – 4bc)^2$
$= (2 \cdot 4 \cdot 6 – 4 \cdot 6 \cdot 3)^2$ [Substituting the values of $a, b$, and $c$]
$= (48 – 72)^2$
$= (-24)^2$
$= 576$(Ans.)
5. If $x – \frac{1}{x} = 3$, find the value of $x^2 + \frac{1}{x^2}$.
Solution: Given, $x – \frac{1}{x} = 3$.
Given expression $= x^2 + \frac{1}{x^2}$
$= \left(x – \frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x}$
$= (3)^2 + 2$
$= 9 + 2$
$= 11$(Ans.)
6. If $a + \frac{1}{a} = 4$, what is the value of $a^4 + \frac{1}{a^4}$?
Solution: Given, $a + \frac{1}{a} = 4$.
Given expression $= a^4 + \frac{1}{a^4}$
$= (a^2)^2 + \left(\frac{1}{a^2}\right)^2$
$= \left(a^2 + \frac{1}{a^2}\right)^2 – 2 \cdot a^2 \cdot \frac{1}{a^2}$
$= \left\{\left(a + \frac{1}{a}\right)^2 – 2 \cdot a \cdot \frac{1}{a}\right\}^2 – 2$
$= \{(4)^2 – 2\}^2 – 2$
$= (16 – 2)^2 – 2$
$= (14)^2 – 2$
$= 196 – 2$
$= 194$(Ans.)
7. If $m = 6, n = 7$, find the value of $16(m^2 + n^2)^2 + 56(m^2 + n^2)(3m^2 – 2n^2) + 49(3m^2 – 2n^2)^2$.
Solution: Given, $m = 6$ and $n = 7$.
Let $m^2 + n^2 = x$ and $3m^2 – 2n^2 = y$.
Given expression $= 16x^2 + 56xy + 49y^2$
$= (4x)^2 + 2 \cdot (4x) \cdot (7y) + (7y)^2$
$= (4x + 7y)^2$
$= \{4(m^2 + n^2) + 7(3m^2 – 2n^2)\}^2$ [Substituting the values of $x$ and $y$]
$= (4m^2 + 4n^2 + 21m^2 – 14n^2)^2$
$= (25m^2 – 10n^2)^2$
$= \{25(6)^2 – 10(7)^2\}^2$ [Substituting the values of $m$ and $n$]
$= \{25(36) – 10(49)\}^2$
$= (900 – 490)^2$
$= (410)^2$
$= 168100$(Ans.)
8. If $a – \frac{1}{a} = m$, show that $a^4 + \frac{1}{a^4} = m^4 + 4m^2 + 2$.
Solution: Given, $a – \frac{1}{a} = m$.
L.H.S. $= a^4 + \frac{1}{a^4}$
$= (a^2)^2 + \left(\frac{1}{a^2}\right)^2$
$= \left(a^2 + \frac{1}{a^2}\right)^2 – 2 \cdot a^2 \cdot \frac{1}{a^2}$
$= \left\{\left(a – \frac{1}{a}\right)^2 + 2 \cdot a \cdot \frac{1}{a}\right\}^2 – 2$
$= (m^2 + 2)^2 – 2$
$= (m^2)^2 + 2 \cdot m^2 \cdot 2 + (2)^2 – 2$
$= m^4 + 4m^2 + 4 – 2$
$= m^4 + 4m^2 + 2$
$= \text{R.H.S.}$
$\therefore a^4 + \frac{1}{a^4} = m^4 + 4m^2 + 2$(Shown)
9. If $x – \frac{1}{x} = 4$, prove that $x^2 + \frac{1}{x^2} = 18$.
Solution: Given, $x – \frac{1}{x} = 4$.
L.H.S. $= x^2 + \frac{1}{x^2}$
$= \left(x – \frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x}$
$= (4)^2 + 2$
$= 16 + 2$
$= 18$
$= \text{R.H.S.}$
$\therefore x^2 + \frac{1}{x^2} = 18$(Proved)
10. If $m + \frac{1}{m} = 2$, prove that $m^4 + \frac{1}{m^4} = 2$.
Solution: Given, $m + \frac{1}{m} = 2$.
L.H.S. $= m^4 + \frac{1}{m^4}$
$= (m^2)^2 + \left(\frac{1}{m^2}\right)^2$
$= \left(m^2 + \frac{1}{m^2}\right)^2 – 2 \cdot m^2 \cdot \frac{1}{m^2}$
$= \left\{\left(m + \frac{1}{m}\right)^2 – 2 \cdot m \cdot \frac{1}{m}\right\}^2 – 2$
$= \{(2)^2 – 2\}^2 – 2$
$= (4 – 2)^2 – 2$
$= (2)^2 – 2$
$= 4 – 2$
$= 2$
$= \text{R.H.S.}$
$\therefore m^4 + \frac{1}{m^4} = 2$(Proved)
11. If $x + y = 12$ and $xy = 27$, find the values of $(x – y)^2$ and $x^2 + y^2$.
Solution:
Given, $x + y = 12$ and $xy = 27$.
First expression:
$(x – y)^2 = (x + y)^2 – 4xy$
$= (12)^2 – 4 \cdot 27$
$= 144 – 108$
$= 36$
Second expression:
$x^2 + y^2 = (x + y)^2 – 2xy$
$= (12)^2 – 2 \cdot 27$
$= 144 – 54$
$= 90$
Answer:$36$ and $90$
12. If $a + b = 13$ and $a – b = 3$, find the values of $2a^2 + 2b^2$ and $ab$.
Solution:
Given, $a + b = 13$ and $a – b = 3$.
First expression:
$2a^2 + 2b^2 = 2(a^2 + b^2)$
$= (a + b)^2 + (a – b)^2$
$= (13)^2 + (3)^2$
$= 169 + 9$
$= 178$
Second expression:
$ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$
$= \left(\frac{13}{2}\right)^2 – \left(\frac{3}{2}\right)^2$
$= \frac{169}{4} – \frac{9}{4}$
$= \frac{169 – 9}{4}$
$= \frac{160}{4}$
$= 40$
Answer:$178$ and $40$
13. Express as the difference of the square of two Expressions:
Necessary Formula:
$ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$
(a) $(5p – 3q)(p + 7q)$
Solution:
Let $5p – 3q = x$ and $p + 7q = y$.
We know, $xy = \left(\frac{x + y}{2}\right)^2 – \left(\frac{x – y}{2}\right)^2$
Substituting the values of $x$ and $y$ we get
$(5p – 3q)(p + 7q)$
$= \left\{\frac{(5p – 3q) + (p + 7q)}{2}\right\}^2 – \left\{\frac{(5p – 3q) – (p + 7q)}{2}\right\}^2$
$= \left(\frac{5p – 3q + p + 7q}{2}\right)^2 – \left(\frac{5p – 3q – p – 7q}{2}\right)^2$
$= \left(\frac{6p + 4q}{2}\right)^2 – \left(\frac{4p – 10q}{2}\right)^2$
$= \left\{\frac{2(3p + 2q)}{2}\right\}^2 – \left\{\frac{2(2p – 5q)}{2}\right\}^2$
$= (3p + 2q)^2 – (2p – 5q)^2$(Ans.)
(b) $(6a + 9b)(7b – 8a)$
Solution:
Let $6a + 9b = x$ and $7b – 8a = y$.
We know, $xy = \left(\frac{x + y}{2}\right)^2 – \left(\frac{x – y}{2}\right)^2$
Substituting the values of $x$ and $y$ we get
$(6a + 9b)(7b – 8a)$
$= \left\{\frac{(6a + 9b) + (7b – 8a)}{2}\right\}^2 – \left\{\frac{(6a + 9b) – (7b – 8a)}{2}\right\}^2$
$= \left(\frac{6a + 9b + 7b – 8a}{2}\right)^2 – \left(\frac{6a + 9b – 7b + 8a}{2}\right)^2$
$= \left(\frac{16b – 2a}{2}\right)^2 – \left(\frac{14a + 2b}{2}\right)^2$
$= \left\{\frac{2(8b – a)}{2}\right\}^2 – \left\{\frac{2(7a + b)}{2}\right\}^2$
$= (8b – a)^2 – (7a + b)^2$(Ans.)
(c) $(3x + 5y)(7x – 5y)$
Solution:
Let $3x + 5y = a$ and $7x – 5y = b$.
We know, $ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$
Substituting the values of $a$ and $b$ we get
$(3x + 5y)(7x – 5y)$
$= \left\{\frac{(3x + 5y) + (7x – 5y)}{2}\right\}^2 – \left\{\frac{(3x + 5y) – (7x – 5y)}{2}\right\}^2$
$= \left(\frac{3x + 5y + 7x – 5y}{2}\right)^2 – \left(\frac{3x + 5y – 7x + 5y}{2}\right)^2$
$= \left(\frac{10x}{2}\right)^2 – \left(\frac{10y – 4x}{2}\right)^2$
$= (5x)^2 – \left\{\frac{2(5y – 2x)}{2}\right\}^2$
$= (5x)^2 – (5y – 2x)^2$(Ans.)
(d) $(5x + 13)(5x – 13)$
Solution:
Let $5x + 13 = a$ and $5x – 13 = b$.
We know, $ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$
Substituting the values of $a$ and $b$ we get
$(5x + 13)(5x – 13)$
$= \left\{\frac{(5x + 13) + (5x – 13)}{2}\right\}^2 – \left\{\frac{(5x + 13) – (5x – 13)}{2}\right\}^2$
$= \left(\frac{5x + 13 + 5x – 13}{2}\right)^2 – \left(\frac{5x + 13 – 5x + 13}{2}\right)^2$
$= \left(\frac{10x}{2}\right)^2 – \left(\frac{26}{2}\right)^2$
$= (5x)^2 – (13)^2$(Ans.)
14. Two numbers are $a$ and $b$, where $a > b$. The sum of the numbers is $12$ and their product is $32$.
Given, $a + b = 12$ and $ab = 32$.
(a) Multiply with the help of formulae: $(2x + 3)(2x – 7)$
Solution:
We know, $(x + a)(x + b) = x^2 + (a + b)x + ab$
$\therefore (2x + 3)(2x – 7)$
$= (2x)^2 + \{3 + (-7)\} \cdot 2x + \{3 \cdot (-7)\}$
$= 4x^2 + (3 – 7) \cdot 2x – 21$
$= 4x^2 + (-4) \cdot 2x – 21$
$= 4x^2 – 8x – 21$(Ans.)
(b) Find the value of $2a^2 + 2b^2$.
Solution:
We know, $(a – b)^2 = (a + b)^2 – 4ab$
$\implies (a – b)^2 = (12)^2 – 4 \cdot 32$
$\implies (a – b)^2 = 144 – 128$
$\implies (a – b)^2 = 16$
Given expression $= 2a^2 + 2b^2$
$= 2(a^2 + b^2)$
$= (a + b)^2 + (a – b)^2$
$= (12)^2 + 16$
$= 144 + 16$
$= 160$(Ans.)
(c) Prove that $(a + 2b)^2 – 5b^2 = 176$.
Solution:
From part (b), we have $(a – b)^2 = 16$.
Since $a > b$, $a – b = \sqrt{16} = 4$.
Now, adding $a + b = 12$ and $a – b = 4$:
$(a + b) + (a – b) = 12 + 4$
$\implies 2a = 16$
$\implies a = 8$
Substituting the value of $a$ into $a + b = 12$:
$8 + b = 12$
$\implies b = 12 – 8$
$\implies b = 4$
L.H.S. $= (a + 2b)^2 – 5b^2$
$= \{8 + 2(4)\}^2 – 5(4)^2$
$= (8 + 8)^2 – 5(16)$
$= (16)^2 – 80$
$= 256 – 80$
$= 176$
$= \text{R.H.S.}$
$\therefore (a + 2b)^2 – 5b^2 = 176$(Proved)