Q46. Without multiplying, evaluate:
(i) $(101)^2$
(ii) $(1002)^2$
(iii) $(502)^2$
(iv) $(97)^2$
(v) $(998)^2$
(vi) $(496)^2$
Show Answer
(i) $(101)^2 = (100 + 1)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (100)^2 + 2(100 \times 1) + (1)^2$
$\Rightarrow 10,000 + 2(100) + 1$
$\Rightarrow 10,000 + 200 + 1$
$\Rightarrow 10,000 + 201 = 10,201$ Ans
(ii) $(1002)^2 = (1000 + 2)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (1000)^2 + 2(1000 \times 2) + (2)^2$
$\Rightarrow 10,00,000 + 2(2000) + 4$
$\Rightarrow 10,00,000 + 2000 + 4$
$\Rightarrow 10,00,000 + 204 = 10,00,204$ Ans
(iii) $(502)^2$
$(502)^2 = (500 + 2)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (500)^2 + 2(500 \times 2) + (2)^2$
$\Rightarrow 2,50,000 + 2(1000) + 4$
$\Rightarrow 2,50,000 + 2000 + 4$
$\Rightarrow 2,50,000 + 2004 = 2,52,004$ Ans
(iv) $(97)^2$
$(97)^2 = (100 - 3)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (100)^2 - 2(100 \times 3) + (3)^2$
$\Rightarrow 10,000 - 2(300) + 9$
$\Rightarrow 10,000 - 600 + 9$
$\Rightarrow 9,400 + 9 = 9,409$ Ans
(v) $(998)^2$
$(998)^2 = (1000 - 2)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (1000)^2 - 2(1000 \times 2) + (2)^2$
$\Rightarrow 10,00,000 - 2(2000) + 4$
$\Rightarrow 10,00,000 - 4000 + 4$
$\Rightarrow 9,96,000 + 4 = 9,96,004$ Ans
(vi) $(496)^2$
$(496)^2 = (500 - 4)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (500)^2 - 2(500 \times 4) + (4)^2$
$\Rightarrow 2,50,000 - 2(2000) + 16$
$\Rightarrow 2,50,000 - 4000 + 16$
$\Rightarrow 2,46,000 + 16 = 2,46,016$ Ans
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (100)^2 + 2(100 \times 1) + (1)^2$
$\Rightarrow 10,000 + 2(100) + 1$
$\Rightarrow 10,000 + 200 + 1$
$\Rightarrow 10,000 + 201 = 10,201$ Ans
(ii) $(1002)^2 = (1000 + 2)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (1000)^2 + 2(1000 \times 2) + (2)^2$
$\Rightarrow 10,00,000 + 2(2000) + 4$
$\Rightarrow 10,00,000 + 2000 + 4$
$\Rightarrow 10,00,000 + 204 = 10,00,204$ Ans
(iii) $(502)^2$
$(502)^2 = (500 + 2)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (500)^2 + 2(500 \times 2) + (2)^2$
$\Rightarrow 2,50,000 + 2(1000) + 4$
$\Rightarrow 2,50,000 + 2000 + 4$
$\Rightarrow 2,50,000 + 2004 = 2,52,004$ Ans
(iv) $(97)^2$
$(97)^2 = (100 - 3)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (100)^2 - 2(100 \times 3) + (3)^2$
$\Rightarrow 10,000 - 2(300) + 9$
$\Rightarrow 10,000 - 600 + 9$
$\Rightarrow 9,400 + 9 = 9,409$ Ans
(v) $(998)^2$
$(998)^2 = (1000 - 2)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (1000)^2 - 2(1000 \times 2) + (2)^2$
$\Rightarrow 10,00,000 - 2(2000) + 4$
$\Rightarrow 10,00,000 - 4000 + 4$
$\Rightarrow 9,96,000 + 4 = 9,96,004$ Ans
(vi) $(496)^2$
$(496)^2 = (500 - 4)^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (500)^2 - 2(500 \times 4) + (4)^2$
$\Rightarrow 2,50,000 - 2(2000) + 16$
$\Rightarrow 2,50,000 - 4000 + 16$
$\Rightarrow 2,46,000 + 16 = 2,46,016$ Ans
Q47. Evaluate:
(i) $(x + 3)(x + 5)$
(ii) $(x - 8)(x + 7)$
(iii) $(x + 4)(x - 1)$
(iv) $(x - 7)(x - 3)$
(v) $(x - 2)(x - 3)(x + 4)$
(vi) $(x - 5)(2x - 1)(2x + 3)$
Show Answer
(i) $(x + 3)(x + 5)$
$\Rightarrow x^2 + (3 + 5)x + (3 \times 5)$
$\Rightarrow x^2 + 8x + 15$ Ans
(ii) $(x - 8)(x + 7)$
$\Rightarrow x^2 - (8 - 7)x - (8 \times 7)$
$\Rightarrow x^2 - x - 56$ Ans
(iii) $(x + 4)(x - 1)$
$\Rightarrow x^2 + (4 - 1)x - (4 \times 1)$
$\Rightarrow x^2 + 3x - 4$ Ans
(iv) $(x - 7)(x - 3)$
$\Rightarrow x^2 - (7 + 3)x + (7 \times 3)$
$\Rightarrow x^2 - 10x + 21$ Ans
(v) $(x - 2)(x - 3)(x + 4)$
$\Rightarrow (x^2 - (4 + 1)x + (3 \times 2))(x + 4)$
$\Rightarrow x(x^2 - 5x + 6) + 4(x^2 - 5x + 6)$
$\Rightarrow x^3 - 5x^2 + 6x + 4x^2 - 20x + 24$
$\Rightarrow x^3 + (4x^2 - 5x^2) + (6x - 20x) + 24$
$\Rightarrow x^3 + (-x^2) + (-14x) + 24$
$\Rightarrow x^3 - x^2 - 14x + 24$ Ans
(vi) $(x - 5)(2x - 1)(2x + 3)$
$\Rightarrow (2x^2 - (10 + 1)x + (5 \times 1))(2x + 3)$
$\Rightarrow (2x^2 - 11x + 5)(2x + 3)$
$\Rightarrow 2x(2x^2 - 11x + 5) + 3(2x^2 - 11x + 5)$
$\Rightarrow 4x^3 - 22x^2 + 10x + 6x^2 - 33x + 15$
$\Rightarrow 4x^3 + (6x^2 - 22x^2) + (10x - 33x) + 15$
$\Rightarrow 4x^3 + (-16x^2) + (-23x) + 15$
$\Rightarrow 4x^3 - 16x^2 - 23x + 15$ Ans
$\Rightarrow x^2 + (3 + 5)x + (3 \times 5)$
$\Rightarrow x^2 + 8x + 15$ Ans
(ii) $(x - 8)(x + 7)$
$\Rightarrow x^2 - (8 - 7)x - (8 \times 7)$
$\Rightarrow x^2 - x - 56$ Ans
(iii) $(x + 4)(x - 1)$
$\Rightarrow x^2 + (4 - 1)x - (4 \times 1)$
$\Rightarrow x^2 + 3x - 4$ Ans
(iv) $(x - 7)(x - 3)$
$\Rightarrow x^2 - (7 + 3)x + (7 \times 3)$
$\Rightarrow x^2 - 10x + 21$ Ans
(v) $(x - 2)(x - 3)(x + 4)$
$\Rightarrow (x^2 - (4 + 1)x + (3 \times 2))(x + 4)$
$\Rightarrow x(x^2 - 5x + 6) + 4(x^2 - 5x + 6)$
$\Rightarrow x^3 - 5x^2 + 6x + 4x^2 - 20x + 24$
$\Rightarrow x^3 + (4x^2 - 5x^2) + (6x - 20x) + 24$
$\Rightarrow x^3 + (-x^2) + (-14x) + 24$
$\Rightarrow x^3 - x^2 - 14x + 24$ Ans
(vi) $(x - 5)(2x - 1)(2x + 3)$
$\Rightarrow (2x^2 - (10 + 1)x + (5 \times 1))(2x + 3)$
$\Rightarrow (2x^2 - 11x + 5)(2x + 3)$
$\Rightarrow 2x(2x^2 - 11x + 5) + 3(2x^2 - 11x + 5)$
$\Rightarrow 4x^3 - 22x^2 + 10x + 6x^2 - 33x + 15$
$\Rightarrow 4x^3 + (6x^2 - 22x^2) + (10x - 33x) + 15$
$\Rightarrow 4x^3 + (-16x^2) + (-23x) + 15$
$\Rightarrow 4x^3 - 16x^2 - 23x + 15$ Ans
Q48. Evaluate:
(i) $(\dfrac{a}{2b} + \dfrac{2b}{a})^2 - (\dfrac{2b}{a} - \dfrac{a}{2b})^2 - 4$
(ii) $(4a + 3b)^2 - (4a - 3b)^2 + 48ab$
Show Answer
(i) $(\dfrac{a}{2b} + \dfrac{2b}{a})^2 - (\dfrac{2b}{a} - \dfrac{a}{2b})^2 - 4$
$\because (a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)$
$\Rightarrow ((\dfrac{a}{2b})^2 + 2(\dfrac{a}{2b} \times \dfrac{2b}{a}) + (\dfrac{2b}{a})^2) - ((\dfrac{2b}{a})^2 - 2(\dfrac{2b}{a} \times \dfrac{a}{2b}) + (\dfrac{a}{2b})^2) - 4$
$\Rightarrow (\dfrac{a^2}{4b^2} + 2(1) + \dfrac{4b^2}{a^2}) - (\dfrac{4b^2}{a^2} - 2(1) + \dfrac{a^2}{4b^2}) - 4$
$\Rightarrow (\dfrac{a^2}{4b^2} + 2 + \dfrac{4b^2}{a^2} - \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2}) - 4$
$\Rightarrow ((\dfrac{a^2}{4b^2} - \dfrac{a^2}{4b^2}) + (\dfrac{4b^2}{a^2} - \dfrac{4b^2}{a^2}) + (2 + 2)) - 4$
$\Rightarrow 4 - 4 = 0$ Ans
(ii) $(4a + 3b)^2 - (4a - 3b)^2 + 48ab$
$\because (a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) + 48ab$
$\Rightarrow ((4a)^2 + 2(4a \times 3b) + (3b)^2) - ((4a)^2 - 2(4a \times 3b) + (3b)^2) + 48ab$
$\Rightarrow (16a^2 + 2(12ab) + 9b^2) - (16a^2 - 2(12ab) + 9b^2) + 48ab$
$\Rightarrow (16a^2 + 24ab + 9b^2) - (16a^2 - 24ab + 9b^2) + 48ab$
$\Rightarrow 16a^2 + 24ab + 9b^2 - 16a^2 + 24ab - 9b^2 + 48ab$
$\Rightarrow (16a^2 - 16a^2) + (24ab + 24ab + 48ab) + (9b^2 - 9b^2)$
$\Rightarrow 96ab$ Ans
$\because (a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)$
$\Rightarrow ((\dfrac{a}{2b})^2 + 2(\dfrac{a}{2b} \times \dfrac{2b}{a}) + (\dfrac{2b}{a})^2) - ((\dfrac{2b}{a})^2 - 2(\dfrac{2b}{a} \times \dfrac{a}{2b}) + (\dfrac{a}{2b})^2) - 4$
$\Rightarrow (\dfrac{a^2}{4b^2} + 2(1) + \dfrac{4b^2}{a^2}) - (\dfrac{4b^2}{a^2} - 2(1) + \dfrac{a^2}{4b^2}) - 4$
$\Rightarrow (\dfrac{a^2}{4b^2} + 2 + \dfrac{4b^2}{a^2} - \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2}) - 4$
$\Rightarrow ((\dfrac{a^2}{4b^2} - \dfrac{a^2}{4b^2}) + (\dfrac{4b^2}{a^2} - \dfrac{4b^2}{a^2}) + (2 + 2)) - 4$
$\Rightarrow 4 - 4 = 0$ Ans
(ii) $(4a + 3b)^2 - (4a - 3b)^2 + 48ab$
$\because (a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) + 48ab$
$\Rightarrow ((4a)^2 + 2(4a \times 3b) + (3b)^2) - ((4a)^2 - 2(4a \times 3b) + (3b)^2) + 48ab$
$\Rightarrow (16a^2 + 2(12ab) + 9b^2) - (16a^2 - 2(12ab) + 9b^2) + 48ab$
$\Rightarrow (16a^2 + 24ab + 9b^2) - (16a^2 - 24ab + 9b^2) + 48ab$
$\Rightarrow 16a^2 + 24ab + 9b^2 - 16a^2 + 24ab - 9b^2 + 48ab$
$\Rightarrow (16a^2 - 16a^2) + (24ab + 24ab + 48ab) + (9b^2 - 9b^2)$
$\Rightarrow 96ab$ Ans
Q49. If $a + b = 7$ and $ab = 10$, find $a - b$
Show Answer
$\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (a - b)^2 = (a + b)^2 - 4ab$
$\Rightarrow (a - b)^2 = (7)^2 - 4(10)$
$\Rightarrow (a - b)^2 = 49 - 40$
$\Rightarrow (a - b)^2 = 9$
$\Rightarrow a - b = \pm\sqrt{9}$
$\Rightarrow a - b = \pm3$ Ans
$\therefore (a - b)^2 = (a + b)^2 - 4ab$
$\Rightarrow (a - b)^2 = (7)^2 - 4(10)$
$\Rightarrow (a - b)^2 = 49 - 40$
$\Rightarrow (a - b)^2 = 9$
$\Rightarrow a - b = \pm\sqrt{9}$
$\Rightarrow a - b = \pm3$ Ans
Q50. If $a - b = 7$ and $ab = 18$, find $a + b$
Show Answer
$\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (a + b)^2 = (a - b)^2 + 4ab$
$\Rightarrow (a + b)^2 = (7)^2 + 4(18)$
$\Rightarrow (a + b)^2 = 49 + 72$
$\Rightarrow (a + b)^2 = 121$
$\Rightarrow a + b = \pm\sqrt{121}$
$\Rightarrow a + b = \pm11$ Ans
$\therefore (a + b)^2 = (a - b)^2 + 4ab$
$\Rightarrow (a + b)^2 = (7)^2 + 4(18)$
$\Rightarrow (a + b)^2 = 49 + 72$
$\Rightarrow (a + b)^2 = 121$
$\Rightarrow a + b = \pm\sqrt{121}$
$\Rightarrow a + b = \pm11$ Ans
Q51. If $x + y = \dfrac{7}{2}$ and $xy = \dfrac{5}{2}$, find:
(i) $x - y$
(ii) $x^2 - y^2$
Show Answer
(i) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (x + y)^2 - (x - y)^2 = 4xy$
$\Rightarrow (x - y)^2 = (x + y)^2 - 4xy$
$\Rightarrow (x - y)^2 = (\dfrac{7}{2})^2 - 4(\dfrac{5}{2})$
$\Rightarrow (x - y)^2 = \dfrac{49}{4} - \dfrac{20}{2}$
$\Rightarrow (x - y)^2 = \dfrac{49}{4} - 10$
$\Rightarrow (x - y)^2 = \dfrac{49 - 40}{4}$
$\Rightarrow (x - y)^2 = \dfrac{9}{4}$
$\Rightarrow a - b = \pm\sqrt{\dfrac{9}{4}}$
$\Rightarrow a - b = \pm\dfrac{3}{2}$ Ans
(ii) $\because$ We have the formula $a^2 - b^2 = (a + b)(a - b)$
Given: $x + y = \dfrac{7}{2}$
From (i) we have: $x - y = \dfrac{3}{2}$
$\therefore x^2 - y^2 = (\dfrac{7}{2})(\dfrac{3}{2})$
$\Rightarrow x^2 - y^2 = \dfrac{7}{2} \times \dfrac{3}{2}$
$\Rightarrow x^2 - y^2 = \pm\dfrac{21}{4}$ Ans
$\therefore (x + y)^2 - (x - y)^2 = 4xy$
$\Rightarrow (x - y)^2 = (x + y)^2 - 4xy$
$\Rightarrow (x - y)^2 = (\dfrac{7}{2})^2 - 4(\dfrac{5}{2})$
$\Rightarrow (x - y)^2 = \dfrac{49}{4} - \dfrac{20}{2}$
$\Rightarrow (x - y)^2 = \dfrac{49}{4} - 10$
$\Rightarrow (x - y)^2 = \dfrac{49 - 40}{4}$
$\Rightarrow (x - y)^2 = \dfrac{9}{4}$
$\Rightarrow a - b = \pm\sqrt{\dfrac{9}{4}}$
$\Rightarrow a - b = \pm\dfrac{3}{2}$ Ans
(ii) $\because$ We have the formula $a^2 - b^2 = (a + b)(a - b)$
Given: $x + y = \dfrac{7}{2}$
From (i) we have: $x - y = \dfrac{3}{2}$
$\therefore x^2 - y^2 = (\dfrac{7}{2})(\dfrac{3}{2})$
$\Rightarrow x^2 - y^2 = \dfrac{7}{2} \times \dfrac{3}{2}$
$\Rightarrow x^2 - y^2 = \pm\dfrac{21}{4}$ Ans
Q52. If $a - b = 0.9$ and $ab = 0.36$, find:
(i) $a + b$
(ii) $a^2 - b^2$
Show Answer
(i) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (a + b)^2 = (a - b)^2 + 4ab$
$\Rightarrow (a + b)^2 = (0.9)^2 + 4(0.36)$
$\Rightarrow (a + b)^2 = 0.81 + 1.44$
$\Rightarrow (a + b)^2 = 2.15$
$\Rightarrow a + b = \pm\sqrt{2.15}$
$\Rightarrow a + b = \pm1.46$ or $\pm1.5$ Ans
(ii) $\because$ We have the formula $a^2 - b^2 = (a + b)(a - b)$
Given: $a - b = 0.9$
From (i) we have: $a + b = 1.5$
$\therefore a^2 - b^2 = (0.9)(1.5)$
$\Rightarrow x^2 - y^2 = 0.9 \times 1.5$
$\Rightarrow x^2 - y^2 = \pm1.35$ Ans
$\therefore (a + b)^2 = (a - b)^2 + 4ab$
$\Rightarrow (a + b)^2 = (0.9)^2 + 4(0.36)$
$\Rightarrow (a + b)^2 = 0.81 + 1.44$
$\Rightarrow (a + b)^2 = 2.15$
$\Rightarrow a + b = \pm\sqrt{2.15}$
$\Rightarrow a + b = \pm1.46$ or $\pm1.5$ Ans
(ii) $\because$ We have the formula $a^2 - b^2 = (a + b)(a - b)$
Given: $a - b = 0.9$
From (i) we have: $a + b = 1.5$
$\therefore a^2 - b^2 = (0.9)(1.5)$
$\Rightarrow x^2 - y^2 = 0.9 \times 1.5$
$\Rightarrow x^2 - y^2 = \pm1.35$ Ans
Q53. If $a + b = 1$ and $a - b = 7$, find:
(i) $a^2 + b^2$
(ii) $ab$
Show Answer
(i) $\because$ We have the formula $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$
$\therefore (1)^2 + (7)^2 = 2(a^2 + b^2)$
$\Rightarrow 1 + 49 = 2(a^2 + b^2)$
$\Rightarrow 2(a^2 + b^2) = 50$
$\Rightarrow a^2 + b^2 = \dfrac{50}{2}$
$\Rightarrow a^2 + b^2 = 25$ Ans
(ii) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (1)^2 - (7)^2 = 4ab$
$\Rightarrow 1 - 49 = 4ab$
$\Rightarrow 4ab = -48$
$\Rightarrow ab = -\dfrac{48}{4}$
$\Rightarrow ab = -12$ Ans
$\therefore (1)^2 + (7)^2 = 2(a^2 + b^2)$
$\Rightarrow 1 + 49 = 2(a^2 + b^2)$
$\Rightarrow 2(a^2 + b^2) = 50$
$\Rightarrow a^2 + b^2 = \dfrac{50}{2}$
$\Rightarrow a^2 + b^2 = 25$ Ans
(ii) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (1)^2 - (7)^2 = 4ab$
$\Rightarrow 1 - 49 = 4ab$
$\Rightarrow 4ab = -48$
$\Rightarrow ab = -\dfrac{48}{4}$
$\Rightarrow ab = -12$ Ans
Q54. If $a - b = 4$ and $a + b = 6$, find:
(i) $a^2 + b^2$
(ii) $ab$
Show Answer
(i) $\because$ We have the formula $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$
$\therefore (6)^2 + (4)^2 = 2(a^2 + b^2)$
$\Rightarrow 36 + 16 = 2(a^2 + b^2)$
$\Rightarrow 2(a^2 + b^2) = 52$
$\Rightarrow a^2 + b^2 = \dfrac{52}{2}$
$\Rightarrow a^2 + b^2 = 26$ Ans
(ii) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (6)^2 - (4)^2 = 4ab$
$\Rightarrow 36 - 16 = 4ab$
$\Rightarrow 4ab = 20$
$\Rightarrow ab = \dfrac{20}{4}$
$\Rightarrow ab = 5$ Ans
$\therefore (6)^2 + (4)^2 = 2(a^2 + b^2)$
$\Rightarrow 36 + 16 = 2(a^2 + b^2)$
$\Rightarrow 2(a^2 + b^2) = 52$
$\Rightarrow a^2 + b^2 = \dfrac{52}{2}$
$\Rightarrow a^2 + b^2 = 26$ Ans
(ii) $\because$ We have the formula $(a + b)^2 - (a - b)^2 = 4ab$
$\therefore (6)^2 - (4)^2 = 4ab$
$\Rightarrow 36 - 16 = 4ab$
$\Rightarrow 4ab = 20$
$\Rightarrow ab = \dfrac{20}{4}$
$\Rightarrow ab = 5$ Ans
Q55. If $a + \dfrac{1}{a} = 6$, find:
(i) $a - \dfrac{1}{a}$
(ii) $a^2 - \dfrac{1}{a^2}$
Show Answer
(i) $\because$ We have the formula $(a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4$
$\therefore (a - \dfrac{1}{a})^2 = (a - \dfrac{1}{a})^2 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = (6)^2 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = 36 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = 32$
$\Rightarrow a - \dfrac{1}{a} = \pm\sqrt{32}$
$\Rightarrow a - \dfrac{1}{a} = \pm\sqrt[4]{2}$ Ans
(ii) $\because a^2 - b^2 = (a + b)(a - b)$
$\therefore a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
Given: $a + \dfrac{1}{a} = 6$
From (i) we have: $a + \dfrac{1}{a} = \sqrt[4]{2}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = (6)(\sqrt[4]{2})$
$\Rightarrow a^2 - \dfrac{1}{a^2} = 6 \times \sqrt[4]{2} = \sqrt[6 \times 4]{2}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm\sqrt[24]{2}$ Ans
$\therefore (a - \dfrac{1}{a})^2 = (a - \dfrac{1}{a})^2 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = (6)^2 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = 36 - 4$
$\Rightarrow (a - \dfrac{1}{a})^2 = 32$
$\Rightarrow a - \dfrac{1}{a} = \pm\sqrt{32}$
$\Rightarrow a - \dfrac{1}{a} = \pm\sqrt[4]{2}$ Ans
(ii) $\because a^2 - b^2 = (a + b)(a - b)$
$\therefore a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
Given: $a + \dfrac{1}{a} = 6$
From (i) we have: $a + \dfrac{1}{a} = \sqrt[4]{2}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = (6)(\sqrt[4]{2})$
$\Rightarrow a^2 - \dfrac{1}{a^2} = 6 \times \sqrt[4]{2} = \sqrt[6 \times 4]{2}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm\sqrt[24]{2}$ Ans
Q56. If $a - \dfrac{1}{a} = 8$, find:
(i) $a + \dfrac{1}{a}$
(ii) $a^2 - \dfrac{1}{a^2}$
Show Answer
(i) $\because$ We have the formula $(a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4$
$\therefore (a + \dfrac{1}{a})^2 = (a - \dfrac{1}{a})^2 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = (8)^2 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 64 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 68$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{68}$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt[2]{17}$ Ans
(ii) $\because a^2 - b^2 = (a + b)(a - b)$
$\therefore a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
Given: $a + \dfrac{1}{a} = 8$
From (i) we have: $a + \dfrac{1}{a} = \sqrt[2]{17}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = (8)(\sqrt[2]{17})$
$\Rightarrow a^2 - \dfrac{1}{a^2} = 8 \times \sqrt[2]{17} = \sqrt[8 \times 2]{17}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm\sqrt[16]{17}$ Ans
$\therefore (a + \dfrac{1}{a})^2 = (a - \dfrac{1}{a})^2 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = (8)^2 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 64 + 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 68$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{68}$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt[2]{17}$ Ans
(ii) $\because a^2 - b^2 = (a + b)(a - b)$
$\therefore a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
Given: $a + \dfrac{1}{a} = 8$
From (i) we have: $a + \dfrac{1}{a} = \sqrt[2]{17}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = (8)(\sqrt[2]{17})$
$\Rightarrow a^2 - \dfrac{1}{a^2} = 8 \times \sqrt[2]{17} = \sqrt[8 \times 2]{17}$
$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm\sqrt[16]{17}$ Ans
Q57. If $a + b + c = 12$ and $a^2 + b^2 + c^2 = 50$, find $ab + bc + ca$
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (12)^2 = 50 + 2(ab + bc + ca)$
$\Rightarrow 144 = 50 + 2(ab + bc + ca)$
$\Rightarrow 144 - 50 = 2(ab + bc + ca)$
$\Rightarrow 2(ab + bc + ca) = 94$
$\Rightarrow ab + bc + ca = \dfrac{94}{2}$
$\Rightarrow ab + bc + ca = 47$ Ans
$\therefore (12)^2 = 50 + 2(ab + bc + ca)$
$\Rightarrow 144 = 50 + 2(ab + bc + ca)$
$\Rightarrow 144 - 50 = 2(ab + bc + ca)$
$\Rightarrow 2(ab + bc + ca) = 94$
$\Rightarrow ab + bc + ca = \dfrac{94}{2}$
$\Rightarrow ab + bc + ca = 47$ Ans
Q58. If $a^2 + b^2 + c^2 = 35$ and $ab + bc + ca = 23$, find $a + b + c$
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (a + b + c)^2 = 35 + 2(23)$
$\Rightarrow (a + b + c)^2 = 35 + 46$
$\Rightarrow (a + b + c)^2 = 81$
$\Rightarrow a + b + c = \sqrt{81}$
$\Rightarrow a + b + c = \pm9$ Ans
$\therefore (a + b + c)^2 = 35 + 2(23)$
$\Rightarrow (a + b + c)^2 = 35 + 46$
$\Rightarrow (a + b + c)^2 = 81$
$\Rightarrow a + b + c = \sqrt{81}$
$\Rightarrow a + b + c = \pm9$ Ans
Q59. If $a + b + c = p$ and $ab + bc + ca = q$, find $a^2 + b^2 + c^2$
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (p)^2 = a^2 + b^2 + c^2 + 2(q)$
$\Rightarrow p^2 = 35 + 2q$
$\Rightarrow p^2 - 2q = 35$ Ans
$\therefore (p)^2 = a^2 + b^2 + c^2 + 2(q)$
$\Rightarrow p^2 = 35 + 2q$
$\Rightarrow p^2 - 2q = 35$ Ans
Q60. If $a^2 - 3a + 1 = 0$, find:
(i) $a + \dfrac{1}{a}$
(ii) $a^2 + \dfrac{1}{a^2}$
Show Answer
(i) Upon dividing each term by $a$ we have:
$\Rightarrow \dfrac{a^2}{a} - \dfrac{3a}{a} + \dfrac{1}{a} = 0$
$\Rightarrow a - 3 + \dfrac{1}{a} = 0$
$\Rightarrow a + \dfrac{1}{a} = 0 + 3$
$\Rightarrow a + \dfrac{1}{a} = 3$ Ans
(ii) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
And from (i) we have: $a + \dfrac{1}{a} = 3$
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 7$ Ans
$\Rightarrow \dfrac{a^2}{a} - \dfrac{3a}{a} + \dfrac{1}{a} = 0$
$\Rightarrow a - 3 + \dfrac{1}{a} = 0$
$\Rightarrow a + \dfrac{1}{a} = 0 + 3$
$\Rightarrow a + \dfrac{1}{a} = 3$ Ans
(ii) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
And from (i) we have: $a + \dfrac{1}{a} = 3$
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 7$ Ans