Q16. If $a - \dfrac{1}{a} = 4$, find: $a^2 + \dfrac{1}{a^2}$
Show Answer
$\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $a - \dfrac{1}{a} = 4$, we have:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 18$ Ans
Given $a - \dfrac{1}{a} = 4$, we have:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 18$ Ans
Q17. If $a^2 + \dfrac{1}{a^2} = 23$, find: $a + \dfrac{1}{a}$
Show Answer
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a^2 + \dfrac{1}{a^2} = 23$, we have:
$\therefore (a + \dfrac{1}{a})^2 = 23 + 2$
$\Rightarrow (a + \dfrac{1}{a})^2 = 25$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{25}$
$\Rightarrow a + \dfrac{1}{a} = \pm5$ Ans
Given $a^2 + \dfrac{1}{a^2} = 23$, we have:
$\therefore (a + \dfrac{1}{a})^2 = 23 + 2$
$\Rightarrow (a + \dfrac{1}{a})^2 = 25$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{25}$
$\Rightarrow a + \dfrac{1}{a} = \pm5$ Ans
Q18. If $a^2 + \dfrac{1}{a^2} = 11$, find: $a - \dfrac{1}{a}$
Show Answer
$\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $a^2 + \dfrac{1}{a^2} = 11$, we have:
$\therefore (a - \dfrac{1}{a})^2 = 11 - 2$
$\Rightarrow (a - \dfrac{1}{a})^2 = 9$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{9}$
$\Rightarrow a + \dfrac{1}{a} = \pm3$ Ans
Given $a^2 + \dfrac{1}{a^2} = 11$, we have:
$\therefore (a - \dfrac{1}{a})^2 = 11 - 2$
$\Rightarrow (a - \dfrac{1}{a})^2 = 9$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{9}$
$\Rightarrow a + \dfrac{1}{a} = \pm3$ Ans
Q19. If $a + b + c = 10$ and $a^2 + b^2 + c^2 = 38$, find: $ab + bc + ca$
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given $a + b + c = 10$ and $a^2 + b^2 + c^2 = 38$, we have:
$\therefore (10)^2 = 38 + 2(ab + bc + ca)$
$\Rightarrow 100 = 38 + 2(ab + bc + ca)$
$\Rightarrow 100 - 38 = 2(ab + bc + ca)$
$\Rightarrow 62 = 2(ab + bc + ca)$
$\Rightarrow ab + bc + ca = \dfrac{62}{2} = 31$ Ans
Given $a + b + c = 10$ and $a^2 + b^2 + c^2 = 38$, we have:
$\therefore (10)^2 = 38 + 2(ab + bc + ca)$
$\Rightarrow 100 = 38 + 2(ab + bc + ca)$
$\Rightarrow 100 - 38 = 2(ab + bc + ca)$
$\Rightarrow 62 = 2(ab + bc + ca)$
$\Rightarrow ab + bc + ca = \dfrac{62}{2} = 31$ Ans
Q20. Find: $a^2 + b^2 + c^2$, if $a + b + c = 9$ and $ab + bc + ca = 24$.
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given $a + b + c = 9$ and $ab + bc + ca = 24$, we have:
$\therefore (9)^2 = a^2 + b^2 + c^2 + 2(24)$
$\Rightarrow 81 = a^2 + b^2 + c^2 + 48$
$\Rightarrow 81 - 48 = a^2 + b^2 + c^2$
$\Rightarrow a^2 + b^2 + c^2 = 33$ Ans
Given $a + b + c = 9$ and $ab + bc + ca = 24$, we have:
$\therefore (9)^2 = a^2 + b^2 + c^2 + 2(24)$
$\Rightarrow 81 = a^2 + b^2 + c^2 + 48$
$\Rightarrow 81 - 48 = a^2 + b^2 + c^2$
$\Rightarrow a^2 + b^2 + c^2 = 33$ Ans
Q21. Find: $a + b + c$, if $a^2 + b^2 + c^2 = 83$ and $ab + bc + ca = 71$.
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given $a^2 + b^2 + c^2 = 83$ and $ab + bc + ca = 71$, we have:
$\therefore (a + b + c)^2 = 83 + 2(71)$
$\Rightarrow (a + b + c)^2 = 83 + 142$
$\Rightarrow (a + b + c)^2 = 225$
$\Rightarrow a + b + c = \pm\sqrt{225}$
$\Rightarrow a + b + c = \pm15$ Ans
Given $a^2 + b^2 + c^2 = 83$ and $ab + bc + ca = 71$, we have:
$\therefore (a + b + c)^2 = 83 + 2(71)$
$\Rightarrow (a + b + c)^2 = 83 + 142$
$\Rightarrow (a + b + c)^2 = 225$
$\Rightarrow a + b + c = \pm\sqrt{225}$
$\Rightarrow a + b + c = \pm15$ Ans
Q22. If $a + b = 6$ and $ab = 8$, find: $a^3 + b^3$.
Show Answer
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
Given $a + b = 6$ and $ab = 8$, we have:
$\therefore (6)^3 = a^3 + b^3 + 3(8)(6)$
$\Rightarrow 216 = a^3 + b^3 + 24(6)$
$\Rightarrow 216 = a^3 + b^3 + 144$
$\Rightarrow 216 - 144 = a^3 + b^3$
$\Rightarrow a^3 + b^3 = 72$ Ans
Given $a + b = 6$ and $ab = 8$, we have:
$\therefore (6)^3 = a^3 + b^3 + 3(8)(6)$
$\Rightarrow 216 = a^3 + b^3 + 24(6)$
$\Rightarrow 216 = a^3 + b^3 + 144$
$\Rightarrow 216 - 144 = a^3 + b^3$
$\Rightarrow a^3 + b^3 = 72$ Ans
Q23. If $a - b = 3$ and $ab = 10$, find: $a^3 - b^3$.
Show Answer
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
Given $a - b = 3$ and $ab = 10$, we have:
$\therefore (3)^3 = a^3 - b^3 - 3(10)(3)$
$\Rightarrow 27 = a^3 - b^3 - 30(3)$
$\Rightarrow 27 = a^3 - b^3 - 90$
$\Rightarrow 27 + 90 = a^3 - b^3$
$\Rightarrow a^3 - b^3 = 117$ Ans
Given $a - b = 3$ and $ab = 10$, we have:
$\therefore (3)^3 = a^3 - b^3 - 3(10)(3)$
$\Rightarrow 27 = a^3 - b^3 - 30(3)$
$\Rightarrow 27 = a^3 - b^3 - 90$
$\Rightarrow 27 + 90 = a^3 - b^3$
$\Rightarrow a^3 - b^3 = 117$ Ans
Q24. Find: $a^3 + \dfrac{1}{a^3}$, if $a + \dfrac{1}{a} = 5$.
Show Answer
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
Given $a + \dfrac{1}{a} = 5$, we have:
$\therefore (5)^3 = a^3 + \dfrac{1}{a^3} + 3(a \times \dfrac{1}{a})(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 3(1)(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 3(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 15$
$\Rightarrow 125 - 15 = a^3 + \dfrac{1}{a^3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 110$ Ans
Given $a + \dfrac{1}{a} = 5$, we have:
$\therefore (5)^3 = a^3 + \dfrac{1}{a^3} + 3(a \times \dfrac{1}{a})(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 3(1)(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 3(5)$
$\Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 15$
$\Rightarrow 125 - 15 = a^3 + \dfrac{1}{a^3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 110$ Ans
Q25. Find: $a^3 - \dfrac{1}{a^3}$, if $a - \dfrac{1}{a} = 4$.
Show Answer
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
Given $a - \dfrac{1}{a} = 4$, we have:
$\therefore (4)^3 = a^3 - \dfrac{1}{a^3} - 3(a \times \dfrac{1}{a})(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 3(1)(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 3(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 12$
$\Rightarrow 64 + 12 = a^3 - \dfrac{1}{a^3}$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 76$ Ans
Given $a - \dfrac{1}{a} = 4$, we have:
$\therefore (4)^3 = a^3 - \dfrac{1}{a^3} - 3(a \times \dfrac{1}{a})(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 3(1)(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 3(4)$
$\Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 12$
$\Rightarrow 64 + 12 = a^3 - \dfrac{1}{a^3}$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 76$ Ans
Q26. If $2x - \dfrac{1}{2x} = 4$, find:
(i) $4x^2 + \dfrac{1}{4x^2}$
(ii) $8x^3 - \dfrac{1}{8x^3}$
Show Answer
(i) $\because (x - \dfrac{1}{x})^2 = x^2 - \dfrac{1}{x^2} - 2$
$\therefore (2x - \dfrac{1}{2x})^2 = 4x^2 + \dfrac{1}{4x^2} - 2$
Given $2x - \dfrac{1}{2x} = 4$, we have:
$\Rightarrow (4)^2 = 4x^2 + \dfrac{1}{4x^2} - 2$
$\Rightarrow 16 = 4x^2 + \dfrac{1}{4x^2} - 2$
$\Rightarrow 16 + 2 = 4x^2 + \dfrac{1}{4x^2}$
$\Rightarrow 4x^2 + \dfrac{1}{4x^2} = 18$ Ans
(ii) $\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (2x - \dfrac{1}{2x})^3 = 8x^3 + \dfrac{1}{8x^3} - 3(2x \times \dfrac{1}{2x})(2x - \dfrac{1}{2x})$
Given $2x - \dfrac{1}{2x} = 4$, we have:
$\Rightarrow (4)^3 = 8x^3 + \dfrac{1}{8x^3} - 3(1)(4)$
$\Rightarrow 64 = 8x^3 + \dfrac{1}{8x^3} - 3(4)$
$\Rightarrow 64 = 8x^3 + \dfrac{1}{8x^3} - 12$
$\Rightarrow 64 + 12 = 8x^3 + \dfrac{1}{8x^3}$
$\Rightarrow 8x^3 + \dfrac{1}{8x^3} = 76$ Ans
$\therefore (2x - \dfrac{1}{2x})^2 = 4x^2 + \dfrac{1}{4x^2} - 2$
Given $2x - \dfrac{1}{2x} = 4$, we have:
$\Rightarrow (4)^2 = 4x^2 + \dfrac{1}{4x^2} - 2$
$\Rightarrow 16 = 4x^2 + \dfrac{1}{4x^2} - 2$
$\Rightarrow 16 + 2 = 4x^2 + \dfrac{1}{4x^2}$
$\Rightarrow 4x^2 + \dfrac{1}{4x^2} = 18$ Ans
(ii) $\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (2x - \dfrac{1}{2x})^3 = 8x^3 + \dfrac{1}{8x^3} - 3(2x \times \dfrac{1}{2x})(2x - \dfrac{1}{2x})$
Given $2x - \dfrac{1}{2x} = 4$, we have:
$\Rightarrow (4)^3 = 8x^3 + \dfrac{1}{8x^3} - 3(1)(4)$
$\Rightarrow 64 = 8x^3 + \dfrac{1}{8x^3} - 3(4)$
$\Rightarrow 64 = 8x^3 + \dfrac{1}{8x^3} - 12$
$\Rightarrow 64 + 12 = 8x^3 + \dfrac{1}{8x^3}$
$\Rightarrow 8x^3 + \dfrac{1}{8x^3} = 76$ Ans
Q27. If $3x + \dfrac{1}{3x} = 3$, find:
(i) $9x^2 + \dfrac{1}{9x^2}$
(ii) $27x^3 + \dfrac{1}{27x^3}$
Show Answer
(i) $\because (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$
$\therefore (3x + \dfrac{1}{3x})^2 = 9x^2 + \dfrac{1}{9x^2} + 2$
Given $3x + \dfrac{1}{3x} = 3$, we have:
$\Rightarrow (3)^2 = 9x^2 + \dfrac{1}{9x^2} + 2$
$\Rightarrow 9 = 4x^2 + \dfrac{1}{4x^2} + 2$
$\Rightarrow 9 - 2 = 9x^2 + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + \dfrac{1}{9x^2} = 7$ Ans
(ii) $\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\therefore (3x + \dfrac{1}{3x})^3 = 27x^3 + \dfrac{1}{27x^3} + 3(3x \times \dfrac{1}{3x})(3x + \dfrac{1}{3x})$
Given $3x + \dfrac{1}{3x} = 3$, we have:
$\Rightarrow (3)^3 = 27x^3 + \dfrac{1}{27x^3} + 3(1)(3)$
$\Rightarrow 27 = 27x^3 + \dfrac{1}{27x^3} + 3(3)$
$\Rightarrow 27 = 27x^3 + \dfrac{1}{27x^3} + 9$
$\Rightarrow 27 - 9 = 27x^3 + \dfrac{1}{27x^3}$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 18$ Ans
$\therefore (3x + \dfrac{1}{3x})^2 = 9x^2 + \dfrac{1}{9x^2} + 2$
Given $3x + \dfrac{1}{3x} = 3$, we have:
$\Rightarrow (3)^2 = 9x^2 + \dfrac{1}{9x^2} + 2$
$\Rightarrow 9 = 4x^2 + \dfrac{1}{4x^2} + 2$
$\Rightarrow 9 - 2 = 9x^2 + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + \dfrac{1}{9x^2} = 7$ Ans
(ii) $\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\therefore (3x + \dfrac{1}{3x})^3 = 27x^3 + \dfrac{1}{27x^3} + 3(3x \times \dfrac{1}{3x})(3x + \dfrac{1}{3x})$
Given $3x + \dfrac{1}{3x} = 3$, we have:
$\Rightarrow (3)^3 = 27x^3 + \dfrac{1}{27x^3} + 3(1)(3)$
$\Rightarrow 27 = 27x^3 + \dfrac{1}{27x^3} + 3(3)$
$\Rightarrow 27 = 27x^3 + \dfrac{1}{27x^3} + 9$
$\Rightarrow 27 - 9 = 27x^3 + \dfrac{1}{27x^3}$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 18$ Ans
Q28. The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) The sum of two numbers.
(ii) The difference between them.
Show Answer
$\Rightarrow$ Let two numbers be $= x$ and $y$
$\therefore x^2 + y^2 = 13$ and $xy$ = 6
(i) $\because (x + y)^2 = x^2 + y^2 + 2xy$
$\therefore (x + y)^2 = 13 + 2 \times 6$
$\Rightarrow (x + y)^2 = 13 + 12$
$\Rightarrow (x + y)^2 = 25$
$\Rightarrow x + y = \pm\sqrt{25}$
$\Rightarrow x + y = \pm5$ Ans
(ii) $\because (x - y)^2 = x^2 + y^2 - 2xy$
$\therefore (x - y)^2 = 13 - 2 \times 6$
$\Rightarrow (x + y)^2 = 13 - 12$
$\Rightarrow (x + y)^2 = 1$
$\Rightarrow x + y = \pm\sqrt{1}$
$\Rightarrow x + y = \pm1$ Ans
$\therefore x^2 + y^2 = 13$ and $xy$ = 6
(i) $\because (x + y)^2 = x^2 + y^2 + 2xy$
$\therefore (x + y)^2 = 13 + 2 \times 6$
$\Rightarrow (x + y)^2 = 13 + 12$
$\Rightarrow (x + y)^2 = 25$
$\Rightarrow x + y = \pm\sqrt{25}$
$\Rightarrow x + y = \pm5$ Ans
(ii) $\because (x - y)^2 = x^2 + y^2 - 2xy$
$\therefore (x - y)^2 = 13 - 2 \times 6$
$\Rightarrow (x + y)^2 = 13 - 12$
$\Rightarrow (x + y)^2 = 1$
$\Rightarrow x + y = \pm\sqrt{1}$
$\Rightarrow x + y = \pm1$ Ans
Q29. Evaluate:
(i) $(3x + \dfrac{1}{2})(2x + \dfrac{1}{3})$
(ii) $(2a + 0.5)(7a - 0.3)$
(iii) $(9 - y)(7 + y)$
(iv) $(2 - z)(15 - z)$
(v) $(a^2 + 5)(a^2 - 3)$
(vi) $(4 - ab)(8 + ab)$
(vii) $(5xy - 7)(7xy + 9)$
(viii) $(3a^2 - 4b^2)(8a^2 - 3b^2)$
Show Answer
(i) $(3x + \dfrac{1}{2})(2x + \dfrac{1}{3})$
$\Rightarrow 3x(2x + \dfrac{1}{3}) + \dfrac{1}{2}(2x + \dfrac{1}{3})$
$\Rightarrow (3x(2x) + 3x(\dfrac{1}{3})) + (\dfrac{1}{2}(2x) + \dfrac{1}{2}(\dfrac{1}{3}))$
$\Rightarrow (6x^2 + \dfrac{3x}{3}) + (\dfrac{2x}{2} + \dfrac{1}{6}))$
$\Rightarrow 6x^2 + x + x + \dfrac{1}{6}$
$\Rightarrow 6x^2 + 2x + \dfrac{1}{6}$ Ans
(ii) $(2a + 0.5)(7a - 0.3)$
$\Rightarrow 2a(7a - 0.3) + 0.5(7a - 0.3)$
$\Rightarrow (2a(7a) - 2a(0.3)) + (0.5(7a) - 0.5(0.3))$
$\Rightarrow (14a^2 - 0.6a) + (3.5a - 0.15)$
$\Rightarrow 14a^2 - 0.6a + 3.5a - 0.15$
$\Rightarrow 14a^2 + 2.9a - 0.15$ Ans
(iii) $(9 - y)(7 + y)$
$\Rightarrow 9(7 + y) - y(7 + y)$
$\Rightarrow (9(7) + 9(y)) - (y(7) - y(y))$
$\Rightarrow (63 + 9y) - (7y - y^2)$
$\Rightarrow 63 + 9y - 7y - y^2$
$\Rightarrow 63 + 2y - y^2$ Ans
(iv) $(2 - z)(15 - z)$
$\Rightarrow 2(15 - z) - z(15 - z)$
$\Rightarrow (2(15) - 2(z)) - (z(15) + z(z))$
$\Rightarrow (30 - 2z) - (15z + z^2)$
$\Rightarrow 30 - 2z - 15z - z^2$
$\Rightarrow 30 - 17z - z^2$ Ans
(v) $(a^2 + 5)(a^2 - 3)$
$\Rightarrow a^2(a^2 - 3) + 5(a^2 - 3)$
$\Rightarrow (a^2(a^2) - a^2(3)) + (5(a^2) - 5(3))$
$\Rightarrow (a^4 - 3a^2) + (5a^2 - 15)$
$\Rightarrow a^4 - 3a^2 + 5a^2 - 15$
$\Rightarrow a^4 + 2a^2 - 15$ Ans
(vi) $(4 - ab)(8 + ab)$
$\Rightarrow 4(8 + ab) - ab(8 + ab)$
$\Rightarrow (4(8) + 4(ab)) - (ab(8) - ab(ab))$
$\Rightarrow (32 + 4ab) - (8ab - ab^2)$
$\Rightarrow 32 + 4ab - 8ab - a^2b^2$
$\Rightarrow 32 - 4ab - a^2b^2$ Ans
(vii) $(5xy - 7)(7xy + 9)$
$\Rightarrow 5xy(7xy + 9) - 7(7xy + 9)$
$\Rightarrow (5xy(7xy) + 5xy(9)) - (7(7xy) - 7(9))$
$\Rightarrow (35x^2y^2 + 45xy) - (49xy - 63)$
$\Rightarrow 35x^2y^2 + 45xy - 49xy - 63$
$\Rightarrow 35x^2y^2 - 4xy - 63$ Ans
(viii) $(3a^2 - 4b^2)(8a^2 - 3b^2)$
$\Rightarrow 3a^2(8a^2 - 3b^2) - 4b^2(8a^2 - 3b^2)$
$\Rightarrow (3a^2(8a^2) - 3a^2(3b^2)) - (4b^2(8a^2) + 4b^2(3b^2))$
$\Rightarrow (24a^4 - 9a^2b^2) - (32a^2b^2 + 12b^4)$
$\Rightarrow 24a^4 - 9a^2b^2 - 32a^2b^2 + 12b^4$
$\Rightarrow 24a^4 - 41a^2b^2 + 12b^4$ Ans
$\Rightarrow 3x(2x + \dfrac{1}{3}) + \dfrac{1}{2}(2x + \dfrac{1}{3})$
$\Rightarrow (3x(2x) + 3x(\dfrac{1}{3})) + (\dfrac{1}{2}(2x) + \dfrac{1}{2}(\dfrac{1}{3}))$
$\Rightarrow (6x^2 + \dfrac{3x}{3}) + (\dfrac{2x}{2} + \dfrac{1}{6}))$
$\Rightarrow 6x^2 + x + x + \dfrac{1}{6}$
$\Rightarrow 6x^2 + 2x + \dfrac{1}{6}$ Ans
(ii) $(2a + 0.5)(7a - 0.3)$
$\Rightarrow 2a(7a - 0.3) + 0.5(7a - 0.3)$
$\Rightarrow (2a(7a) - 2a(0.3)) + (0.5(7a) - 0.5(0.3))$
$\Rightarrow (14a^2 - 0.6a) + (3.5a - 0.15)$
$\Rightarrow 14a^2 - 0.6a + 3.5a - 0.15$
$\Rightarrow 14a^2 + 2.9a - 0.15$ Ans
(iii) $(9 - y)(7 + y)$
$\Rightarrow 9(7 + y) - y(7 + y)$
$\Rightarrow (9(7) + 9(y)) - (y(7) - y(y))$
$\Rightarrow (63 + 9y) - (7y - y^2)$
$\Rightarrow 63 + 9y - 7y - y^2$
$\Rightarrow 63 + 2y - y^2$ Ans
(iv) $(2 - z)(15 - z)$
$\Rightarrow 2(15 - z) - z(15 - z)$
$\Rightarrow (2(15) - 2(z)) - (z(15) + z(z))$
$\Rightarrow (30 - 2z) - (15z + z^2)$
$\Rightarrow 30 - 2z - 15z - z^2$
$\Rightarrow 30 - 17z - z^2$ Ans
(v) $(a^2 + 5)(a^2 - 3)$
$\Rightarrow a^2(a^2 - 3) + 5(a^2 - 3)$
$\Rightarrow (a^2(a^2) - a^2(3)) + (5(a^2) - 5(3))$
$\Rightarrow (a^4 - 3a^2) + (5a^2 - 15)$
$\Rightarrow a^4 - 3a^2 + 5a^2 - 15$
$\Rightarrow a^4 + 2a^2 - 15$ Ans
(vi) $(4 - ab)(8 + ab)$
$\Rightarrow 4(8 + ab) - ab(8 + ab)$
$\Rightarrow (4(8) + 4(ab)) - (ab(8) - ab(ab))$
$\Rightarrow (32 + 4ab) - (8ab - ab^2)$
$\Rightarrow 32 + 4ab - 8ab - a^2b^2$
$\Rightarrow 32 - 4ab - a^2b^2$ Ans
(vii) $(5xy - 7)(7xy + 9)$
$\Rightarrow 5xy(7xy + 9) - 7(7xy + 9)$
$\Rightarrow (5xy(7xy) + 5xy(9)) - (7(7xy) - 7(9))$
$\Rightarrow (35x^2y^2 + 45xy) - (49xy - 63)$
$\Rightarrow 35x^2y^2 + 45xy - 49xy - 63$
$\Rightarrow 35x^2y^2 - 4xy - 63$ Ans
(viii) $(3a^2 - 4b^2)(8a^2 - 3b^2)$
$\Rightarrow 3a^2(8a^2 - 3b^2) - 4b^2(8a^2 - 3b^2)$
$\Rightarrow (3a^2(8a^2) - 3a^2(3b^2)) - (4b^2(8a^2) + 4b^2(3b^2))$
$\Rightarrow (24a^4 - 9a^2b^2) - (32a^2b^2 + 12b^4)$
$\Rightarrow 24a^4 - 9a^2b^2 - 32a^2b^2 + 12b^4$
$\Rightarrow 24a^4 - 41a^2b^2 + 12b^4$ Ans
Q30. Evaluate:
(i) $(2x - \dfrac{3}{5})(2x + \dfrac{3}{5})$
(ii) $(\dfrac{4}{7}a + \dfrac{3}{4}b)(\dfrac{4}{7}a - \dfrac{3}{4}b)$
(iii) $(6 - 5xy)(6 + 5xy)$
(iv) $(2a + \dfrac{1}{2a})(2a - \dfrac{1}{2a})$
(v) $(4x^2 - 5y^2)(4x^2 + 5y^2)$
(vi) $(1.6x + 0.7y)(1.6x - 0.7y)$
(vii) $(m + 3)(m - 3)(m^2 + 9)$
(viii) $(3x + 4y)(3x - 4y)(9x^2 + 16y^2)$
(ix) $(a + bc)(a - bc)(a^2 + b^2c^2)$
Show Answer
(i) $(2x - \dfrac{3}{5})(2x + \dfrac{3}{5})$
$\Rightarrow (2x)^2 - (\dfrac{3}{5})^2$
$\Rightarrow 4x^2 - \dfrac{9}{25}$ Ans
(ii) $(\dfrac{4}{7}a + \dfrac{3}{4}b)(\dfrac{4}{7}a - \dfrac{3}{4}b)$
$\Rightarrow (\dfrac{4}{7}a)^2 - (\dfrac{3}{4}b)^2$
$\Rightarrow \dfrac{16}{49}a^2 - \dfrac{9}{16}b^2$ Ans
(iii) $(6 - 5xy)(6 + 5xy)$
$\Rightarrow (6)^2 - (5xy)^2$
$\Rightarrow 36 - 25x^2y^2$ Ans
(iv) $(2a + \dfrac{1}{2a})(2a - \dfrac{1}{2a})$
$\Rightarrow (2a)^2 - (\dfrac{1}{2a})^2$
$\Rightarrow 4a^2 - \dfrac{1}{4a^2}$ Ans
(v) $(4x^2 - 5y^2)(4x^2 + 5y^2)$
$\Rightarrow (4x^2)^2 - (5y^2)^2$
$\Rightarrow 16x^4 - 25y^4$ Ans
(vi) $(1.6x + 0.7y)(1.6x - 0.7y)$
$\Rightarrow (1.6x)^2 - (0.7y)^2$
$\Rightarrow 2.56x^2 - 0.49y^2$ Ans
(vii) $(m + 3)(m - 3)(m^2 + 9)$
$\Rightarrow (m)^2 - (3)^2 (m^2 + 9)$
$\Rightarrow (m^2 - 9) (m^2 + 9)$
$\Rightarrow (m^2)^2 - (9)^2$
$\Rightarrow m^4 - 81$ Ans
(viii) $(3x + 4y)(3x - 4y)(9x^2 + 16y^2)$
$\Rightarrow (3x)^2 - (4y)^2 (9x^2 + 16y^2)$
$\Rightarrow (9x^2 - 16y^2) (9x^2 + 16y^2)$
$\Rightarrow (9x^2)^2 - (16y^2)^2$
$\Rightarrow 81x^4 - 256y^4$ Ans
(ix) $(a + bc)(a - bc)(a^2 + b^2c^2)$
$\Rightarrow (a)^2 - (bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2) (a^2 + b^2c^2)$
$\Rightarrow (a^2)^2 - (b^2c^2)^2$
$\Rightarrow a^4 - b^4c^4$ Ans
$\Rightarrow (2x)^2 - (\dfrac{3}{5})^2$
$\Rightarrow 4x^2 - \dfrac{9}{25}$ Ans
(ii) $(\dfrac{4}{7}a + \dfrac{3}{4}b)(\dfrac{4}{7}a - \dfrac{3}{4}b)$
$\Rightarrow (\dfrac{4}{7}a)^2 - (\dfrac{3}{4}b)^2$
$\Rightarrow \dfrac{16}{49}a^2 - \dfrac{9}{16}b^2$ Ans
(iii) $(6 - 5xy)(6 + 5xy)$
$\Rightarrow (6)^2 - (5xy)^2$
$\Rightarrow 36 - 25x^2y^2$ Ans
(iv) $(2a + \dfrac{1}{2a})(2a - \dfrac{1}{2a})$
$\Rightarrow (2a)^2 - (\dfrac{1}{2a})^2$
$\Rightarrow 4a^2 - \dfrac{1}{4a^2}$ Ans
(v) $(4x^2 - 5y^2)(4x^2 + 5y^2)$
$\Rightarrow (4x^2)^2 - (5y^2)^2$
$\Rightarrow 16x^4 - 25y^4$ Ans
(vi) $(1.6x + 0.7y)(1.6x - 0.7y)$
$\Rightarrow (1.6x)^2 - (0.7y)^2$
$\Rightarrow 2.56x^2 - 0.49y^2$ Ans
(vii) $(m + 3)(m - 3)(m^2 + 9)$
$\Rightarrow (m)^2 - (3)^2 (m^2 + 9)$
$\Rightarrow (m^2 - 9) (m^2 + 9)$
$\Rightarrow (m^2)^2 - (9)^2$
$\Rightarrow m^4 - 81$ Ans
(viii) $(3x + 4y)(3x - 4y)(9x^2 + 16y^2)$
$\Rightarrow (3x)^2 - (4y)^2 (9x^2 + 16y^2)$
$\Rightarrow (9x^2 - 16y^2) (9x^2 + 16y^2)$
$\Rightarrow (9x^2)^2 - (16y^2)^2$
$\Rightarrow 81x^4 - 256y^4$ Ans
(ix) $(a + bc)(a - bc)(a^2 + b^2c^2)$
$\Rightarrow (a)^2 - (bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2) (a^2 + b^2c^2)$
$\Rightarrow (a^2)^2 - (b^2c^2)^2$
$\Rightarrow a^4 - b^4c^4$ Ans