 ## Class 10 NCERT

### NCERT

1   Check whether the following are quadratic equations : $\\$ (i) $(x + 1)^2 = 2(x – 3)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (ii) $x^2 – 2x = (–2) (3 – x)$ $\\$ (iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$ $\ \ \ \ \ \ \ \$ (iv) $(x – 3)(2x +1) = x(x + 5)$ $\\$ (v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$ $\ \ \ \ \ \ \ \ \$ (vi) $x^2 + 3x + 1 = (x – 2)^2$ $\\$ (vii) $(x + 2)^3 = 2x (x2 – 1)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$

##### Solution :

(i) $(x + 1)^2 = 2(x - 3) \Rightarrow x^2 + 2x + 1 = 2x - 6 \Rightarrow x^2 + 7 = 0$ $\\$ It is of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (ii) $x^2 - 2x = (-2)(3 - x) \Rightarrow x^2 - 2x = - 6 + 2x \Rightarrow x^2 - 4x + 6 = 0$ $\\$ It is not of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (iii) $(x – 2)(x + 1) = (x – 1)(x + 3) \Rightarrow x^2 - x - 2 = x^2 + 2x - 3 \Rightarrow 3x -1 = 0$ $\\$ It is of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (iv) $(x – 3)(2x +1) = x(x + 5) \Rightarrow 2x^2 - 5x - 3 = x^2 + 5x \Rightarrow x^2 - 10x - 3 = 0$ $\\$ It is of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (v) $(2x – 1)(x – 3) = (x + 5)(x – 1) \Rightarrow 2x^2 - 7x +3 = x^2 + 4x - 5 \Rightarrow x^2 - 11x + 8 = 0$ $\\$ It is of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (vi) $x^2 + 3x + 1 = (x – 2)^2 \Rightarrow x^2 + 3x + 1 = x^2 + 4 - 4x \Rightarrow 7x - 3 = 0$ $\\$ It is not of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (vii) $(x + 2)^3 = 2x (x^2 – 1) \Rightarrow x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x \Rightarrow x^3 - 14x - 6x^2 - 8 = 0$ $\\$ It is not of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$ (viii) $x^3 – 4x^3 – x + 1 = (x – 2)^3 \Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x^2 + 12x \Rightarrow 2 x^2 -13x + 9 = 0$ $\\$ It is of the form $ax^2 + bx + c = 0$ $\\$ Hence, the given equation is a quadratic equation. $\\$

2   Represent the following situations in the form of quadratic equations : $\\$ (i) The area of a rectangular plot is $528\ m^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. $\\$ (ii) The product of two consecutive positive integers is $306$. We need to find the integers. $\\$ (iii) Rohan’s mother is $26$ years older than him. The product of their ages (in years) $3$ years from now will be $360$. We would like to find Rohan’s present age. $\\$ (iv) A train travels a distance of $480\ km$ at a uniform speed. If the speed had been $8\ km/h$ less, then it would have taken $3\ hours$ more to cover the same distance. We need to find the speed of the train.

##### Solution :

(ii) Let the consecutive integers be $x$ and $x + 1$. $\\$ It is given that their product is $306.$ $\\$ $\therefore x(x + 1) = 306 \Rightarrow x^2 + x - 306 = 0$

(iv) Let the speed of train be $x\ km/h.$ $\\$ Time taken to travel $480\ km = \dfrac{480}{x}hrs$ $\\$ In second condition, let the speed of train $= (x - 8)\ km/h$ $\\$ It is also given that the train will take $3$ hours to cover the same distance. $\\$ Therefore, time taken to travel $480 km = \Big(\dfrac{480}{x} + 3\Big)hrs$ $\\$ $(x - 8)\Big(\dfrac{480}{x} + 3\Big) = 480$ $\\$ $\Rightarrow 480 + 3x - \dfrac{3840}{x} - 24 = 480$ $\\$ $\Rightarrow 3x - \dfrac{3480}{x} - 24 = 480$ $\\$ $\Rightarrow 3x^2 - 24x + 3840 = 0$ $\\$ $\Rightarrow x^2 - 8x + 1280 = 0$ $\\$

(i) Let the breadth of the plot be $x\ m$. $\\$ Hence, the length of the plot is $(2x + 1)\ m$. $\\$ Area of a rectangle = Length × Breadth $\\$ $\therefore 528 = x (2x + 1)$ $\\$ $\Rightarrow 2x^2 + x - 528 = 0$

(iii) Let Rohan’s age be $x$. $\\$ Hence, his mother’s age $= x + 26$ $3$ years hence, Rohan’s age $= x + 3$ $\\$ Mother’s age $= x + 26 + 3 = x + 29$ $\\$ It is given that the product of their ages after $3$ years is $360.$ $\\$ $\therefore (x + 3)(x + 29) = 360$ $\\$ $\Rightarrow x^2 + 32 x - 273 = 0$

3   Find the roots of the following quadratic equations by factorisation: $\\$ (i) $x^2 – 3x – 10 = 0$ $\\$ (ii) $2x^2 + x – 6 = 0$ $\\$ (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ $\\$ (iv) $2x^2 – x + \dfrac{1}{8}= 0$ $\\$ (v) $100 x^2 – 20x + 1 = 0$

##### Solution :

(i) $x^2 - 3x -10$ $\\$ $= x^2 - 5x + 2x -10$ $\\$ $= x(x - 5) + 2 (x - 5)$ $\\$ $= (x - 5)(x + 2)$ $\\$ Roots of this equation are the values for which $(x -5)(x + 2) = 0$ $\\$ $\therefore x -5 = 0$ or $x + 2 = 0$ $\\$ i.e., $x = 5$ or $x = - 2$

(ii) $2x^2 + x - 6$ $\\$ $= 2x^2 + 4x - 3x - 6$ $\\$ $= 2x(x + 2) - 3(x + 2)$ $\\$ $= (x + 2)(2x - 3)$ $\\$ Roots of this equation are the values for which $(x + 2)(2x - 3) = 0$ $\\$ $\therefore x + 2 = 2x -3 = 0$ $\\$ i.e., $x = -2$ or $x = \dfrac{3}{2}$

(iii) $\sqrt{2}x^2 + 7x + 5 \sqrt{2}$ $\\$ $= \sqrt{2}x^2 + 5x + 2x + 5 \sqrt{2}$ $\\$ $= x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5)$ $\\$ $= (\sqrt{2}x + 5)(x + \sqrt{2})$ $\\$ Roots of this equation are the values for which $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$ $\\$ $\therefore \sqrt{2}x + 5 = 0$ or $x + \sqrt{2} = 0$ $\\$ i.e., $x = \dfrac{-5}{\sqrt{2}}$ or $x = -\sqrt{2}$

(iv) $2x^2 - x + \dfrac{1}{8}$ $\\$ $= \dfrac{1}{8}(16x^3 - 8x + 1)$ $\\$ $= \dfrac{1}{8}(16x^3 - 4x - 4x + 1)$ $\\$ $= \dfrac{1}{8}(4x(4x - 1) - 1 (4x - 1))$ $\\$ $= \dfrac{1}{8}(4x - 1)^2$ $\\$ Roots of this equation are the values for which $(4x -1)^2 = 0$ $\\$ Therefore, $(4x - 1) = 0$ or $(4x - 1) = 0$ $\\$ i.e., $x = \dfrac{1}{4}$ or $x = \dfrac{1}{4}$

(v) $100x^2 - 20x + 1$ $\\$ $= 100x^2 - 10x - 10x + 1$ $\\$ $= 10x (10x - 1) - 1(10x - 1)$ $\\$ $= (10x - 1)^2$ $\\$ Roots of this equation are the values for which $(10x -1)^2 = 0$ $\\$ Therefore, $(10x - 1) = 0$ or $(10x -1) = 0$ $\\$ i.e., $x = \dfrac{1}{10}$ or $x = \dfrac{1}{10}$

4   Solve the problems given in Example $1$.

##### Solution :

$\therefore (55 - x) = 750$ $\\$ $\Rightarrow x^2 - 55x + 750 = 0$ $\\$ $\Rightarrow - 25x - 30x + 750 = 0$ $\\$ $\Rightarrow x (x - 25) - 30 (x - 25) = 0$ $\\$ $\Rightarrow (x - 25)(x -30) = 0$ $\\$ Either $x - 25 = 0$ or $x - 30 = 0$ $\\$ i.e., $x = 25$ or $x = 30$ $\\$ Hence, the number of toys will be either $25$ or $30.$

(i) Let the number of John’s marbles be $x.$ $\\$ Therefore, number of Jivanti’s marble $= 45 - x$ $\\$ After losing $5$ marbles, $\\$ Number of John’s marbles $= x - 5$ $\\$ Number of Jivanti’s marbles $= 45 - x - 5 = 40 - x$ $\\$ It is given that the product of their marbles is $124.$ $\\$ $\therefore (x - 5)(40 - x) = 124$ $\\$ $\Rightarrow x^2 - 45x + 324 = 0$ $\\$ $\Rightarrow x^2 - 36x - 9x + 324$ $\\$ $\Rightarrow x (x - 36) - 9(x -36) = 0$ $\\$ $\Rightarrow (x -36)(x - 9) = 0$ $\\$ Either $x - 36 = 0$ or $x - 9 = 0$ $\\$ i.e., $x = 36$ or $x = 9$ $\\$ If the number of John’s marbles $= 36$, $\\$ Then, number of Jivanti’s marbles $= 45 - 36 = 9$ $\\$ If number of John’s marbles $= 9$, $\\$ Then, number of Jivanti’s marbles $= 45 - 9 = 36$ $\\$ (ii) Let the number of toys produced be $x$. $\\$ $\therefore$ Cost of production of each toy $= Rs\ (55 - x)$ $\\$ It is given that, total production of the toys $= Rs\ 750$

5   Find two numbers whose sum is $27$ and product is $182.$

##### Solution :

Let the first number be $x$ and the second number is $27 - x.$ $\\$ Therefore, their product $= x (27 - x)$ $\\$ It is given that the product of these numbers is $182.$ $\\$ Therefore, $x (27 - x) = 182$ $\\$ $\Rightarrow x^2 - 27x + 182 = 0$ $\\$ $\Rightarrow x^2 - 13x - 14x + 182 = 0$ $\\$ $\Rightarrow x(x - 13) - 14 (x + 13) = 0$ $\\$ $\Rightarrow (x - 13)(x - 14) = 0$ $\\$ Either $x - 13 = 0$ or $x - 14 = 0$ $\\$ i.e., $x = 13$ or $x = 14$ $\\$ If first number $= 13$, then $\\$ Other number $= 27 - 13 = 14$ $\\$ If first number $= 14$, then $\\$ Other number $= 27 - 14 = 13$ $\\$ Therefore, the numbers are $13$ and $14.$

6   Find two consecutive positive integers, sum of whose squares is $365.$

##### Solution :

Let the consecutive positive integers be $x$ and $x + 1.$ $\\$ Given that, $x^2 + (x + 1)^2 = 365$ $\\$ $\Rightarrow x^2 + x^2 + 1 + 2x = 365$ $\\$ $\Rightarrow 2x^2 + 2x - 365 = 0$ $\\$ $\Rightarrow x^2 + x - 182 = 0$ $\\$ $\Rightarrow x^2 + 14x - 13x - 182 = 0$ $\\$ $\Rightarrow x(x + 14) - 13(x + 14) = 0$ $\\$ $\Rightarrow (x + 14)(x - 13) = 0$ $\\$ Either $x + 14 = 0$ or $x - 13 = 0,$ $\\$ i.e., $x = -14$ or $x = 13$ $\\$ Since the integers are positive, $x$ can only be $13$. $\\$ $\therefore x + 1 = 13 + 1 = 14$ $\\$ Therefore, two consecutive positive integers will be $13$ and $14.$

7   The altitude of a right triangle is $7\ cm$ less than its base. If the hypotenuse is $13\ cm$, find the other two sides.

##### Solution :

Let the base of the right triangle be $x\ cm.$ $\\$ Its altitude $= (x - 7)\ cm$ $\\$ From pythagoras theorem, $\\$ $Base^2\ +\ Altitude^2\ =\ Hypotenuse^2$ $\\$ $\therefore x^2 + (x - 7)^2 =13^2$ $\\$ $\Rightarrow x^2 + x^2 + 49 - 14x = 169$ $\\$ $\Rightarrow 2x^2 - 14x - 120 = 0$ $\\$ $\Rightarrow x^2 - 7x - 60 = 0$ $\\$ $\Rightarrow x^2 + 12x + 5x - 60 = 0$ $\\$ $\Rightarrow x(x - 12)+ 5(x - 12) = 0$ $\\$ $\Rightarrow (x - 12)(x - 12) = 0$ $\\$ Either $x - 12 = 0$ or $x + 5 = 0$, i.e., $x = 12$ or $x = -5$ $\\$ Since sides are positive, $x$ can only be $12.$ $\\$ Therefore, the base of the given triangle is $12\ cm$ and the altitude of this triangle will be $(12 - 7)\ cm = 5\ cm.$

8   A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was $Rs\ 90$, find the number of articles produced and the cost of each article.

##### Solution :

Let the number of articles produced be $x.$ $\\$ Therefore, cost of production of each article $= Rs (2x + 3)$ $\\$ It is given that the total production is $Rs\ 90.$ $\\$ $\therefore x(2x + 3) = 90$ $\\$ $\Rightarrow 2x^2 + 3x - 90 = 0$ $\\$ $\Rightarrow 2x^2 + 15x - 12x - 90 = 0$ $\\$ $\Rightarrow x(2x + 15) - 6(2x + 15) = 0$ $\\$ $\Rightarrow (2x + 15)(x - 6) = 0$ $\\$ Either $2x + 15 = 0$ or $x - 6 = 0,$ i.e., $x = \dfrac{-15}{2}$ or $x = 6$ $\\$ As the number of articles produced can only be a positive integer, $\\$ therefore, $x$ can only be $6$. $\\$ Hence, number of articles produced $= 6$ $\\$ Cost of each article $= 2 \times 6 + 3 = Rs 15$

9   Find the roots of the following quadratic equations, if they exist, by the method of completing the square: $\\$ (i) $2x^2 – 7x + 3 = 0$ $\\$ (ii) $2x^2 + x – 4 = 0$ $\\$ (iii) $4x^2 + 4\sqrt{3x} + 3 = 0$ $\\$ (iv) $2x^2 + x + 4 = 0$

##### Solution :

(ii) $2x^2 + x - 4 = 0$ $\\$ $\Rightarrow 2x^2 + x = 4$ $\\$ On dividing both sides of the equation by 2, we obtain $\\$ $\Rightarrow x^2 + \dfrac{1}{2}x = 2$ $\\$ On adding $\Big(\dfrac{1}{4}\Big)^2$ to both sides of the equation, we obtain $\\$ $\Rightarrow (x)^2 + 2 \times x \times \dfrac{1}{4} + \Big(\dfrac{1}{4}\Big)^2 = 2 + \Big(\dfrac{1}{4}\Big)^2$ $\\$ $\Rightarrow \Big(x + \dfrac{1}{4}\Big)^2 = \dfrac{33}{16}$ $\\$ $\Rightarrow x + \dfrac{1}{4} = \pm \dfrac{\sqrt{33}}{4}$ $\\$ $\Rightarrow x = \pm \dfrac{\sqrt{33}}{4} - \dfrac{1}{4}$ $\\$ $\Rightarrow x = \dfrac{\pm \sqrt{33} -1}{4}$ $\\$ $x = \dfrac{\sqrt{33} - 1}{4}$ or $\dfrac{-\sqrt{33} - 1}{4}$ $\\$

(iv) $2x^2 + x + 4 = 0$ $\\$ $\Rightarrow 2x^2 + x = -4$ $\\$ On dividing both sides of the equation by 2, we obtain $\\$ $\Rightarrow x^2 + \dfrac{}{} x = -2$ $\\$ $\Rightarrow x^2 + 2 \times x \times \dfrac{1}{4} = -2$ $\\$ On adding $\Big(\dfrac{1}{4}\Big)^2$ to both sides of the equation, we obtain $\\$ $\Rightarrow (x)^2 + 2 \times x \times \dfrac{1}{4} + \Big(\dfrac{1}{4}\Big)^2 - 2$ $\\$ $\Rightarrow \Big(x + \dfrac{1}{4}\Big)^2 = \dfrac{1}{16} - 2$ $\\$ $\Rightarrow \Big(x + \dfrac{1}{4}\Big)^2 = - \dfrac{31}{16}$ $\\$ However, the square of a number cannot be negative. $\\$ Therefore, there is no real root foe the given equation.

(i) $2x^2 - 7x + 3 = 0$ $\\$ $\Rightarrow 2x^2 - 7x = -3$ $\\$ on dividing both sides of the equation by 2, we obtain $\\$ $\Rightarrow x^2 - \dfrac{7}{2}x = -\dfrac{3}{2}$ $\\$ $\Rightarrow x^2 - 2 \times x \times \dfrac{7}{4} = -\dfrac{3}{2}$ $\\$ On adding $\Big(\dfrac{7}{4}\Big)^2$ to both sides of equation, we obtain $\\$ $\Rightarrow (x)^2 - 2 \times x \times \dfrac{7}{4} + \Big(\dfrac{7}{4}\Big)^2 = \Big(\dfrac{7}{4}\Big)^2 - \dfrac{3}{2}$ $\\$ $\Rightarrow \Big(x - \dfrac{7}{4}\Big)^2 = \dfrac{49}{16} - \dfrac{3}{2}$ $\\$ $\Rightarrow \Big(x - \dfrac{7}{4}\Big)^2 = \dfrac{25}{16}$ $\\$ $\Rightarrow \Big(x - \dfrac{7}{4}\Big) = \pm \dfrac{5}{4}$ $\\$ $\Rightarrow x =\dfrac{7}{4} + \dfrac{5}{4}$ or $x = \dfrac{7}{4} - \dfrac{5}{4}$ $\\$ $\Rightarrow x = \dfrac{12}{4}$ or $x = \dfrac{2}{4}$ $\\$ $\Rightarrow x = 3$ or $\dfrac{1}{2}$ $\\$

(iii) $4x^2 + 4\sqrt{3}x + 3 = 0$ $\\$ $\Rightarrow (2x)^2 + 2 \times 2x \times \sqrt{3} + (\sqrt{3})^2 = 0$ $\\$ $\Rightarrow (2x + \sqrt{3})^2 = 0$ $\\$ $\Rightarrow (2x + \sqrt{3}) = 0$ and $(2x + \sqrt{3}) = 0$ $\\$ $\Rightarrow x = \dfrac{-\sqrt{3}}{2}$ and $x = \dfrac{-\sqrt{3}}{2}$ $\\$

10   Find the roots of the quadratic equations given in $Q.1$ above by applying the quadratic formula. $\\$

##### Solution :

(ii) $2x^2 + x - 4 = 0$ $\\$ On comparing this equation with $ax^2 + bx + c = 0,$ we obtain $\\$ $a = 2,\ b = 1, c = - 4$ $\\$ By using quadratic formula, we obtain $\\$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $\Rightarrow x = \dfrac{-1 \pm \sqrt{1 + 32}}{4}$ $\\$ $\Rightarrow x = \dfrac{-1 \pm \sqrt{33}}{4}$ $\\$ $\therefore x = \dfrac{-1 + \sqrt{33}}{4}$ or $\dfrac{-1 - \sqrt{33}}{4}$ $\\$

(iv) $2x^2 + x + 4 = 0$ $\\$ On comparing this equation with $ax^2 + bx + c = 0$ $\\$ $a = 2,\ b = 1,\ c = 4$ $\\$ By using quadratic formula, we obtain $\\$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $\Rightarrow x = \dfrac{ -1 \pm \sqrt{1 - 32}}{4}$ $\\$ $\Rightarrow x = \dfrac{ -1 \pm \sqrt{31}}{4}$ $\\$ However, the square of a number connot be negative. $\\$ Therefore, there is no real root for the given equation.

(i) $2x^2 - 7x + 3 = 0$ $\\$ On comparing this equation with $ax^2 + bc + c = 0$, we obtian $\\$ $a = 2,\ b = -7,\ c = 3$ $\\$ By using quadratic formula, we obtain $\\$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $\Rightarrow x = \dfrac{7 \pm \sqrt{49 - 24}}{4}$ $\\$ $\Rightarrow x = \dfrac{7 \pm \sqrt{25}}{4}$ $\\$ $\Rightarrow x = \dfrac{7 \pm 5}{4}$ $\\$ $\Rightarrow x = \dfrac{7 + 5}{4}$ or $\dfrac{7 - 5}{4}$ $\\$ $\Rightarrow x = \dfrac{12}{4}$ or $\dfrac{2}{4}$ $\\$ $\therefore x = 3$ or $\dfrac{1}{2}$ $\\$

(iii) $4x^2 + 4\sqrt{3}x + 3 =0$ $\\$ On comparing this equation with $ax^2 + bx + c = 0$ $\\$ $a = 4,\ b = 4\sqrt{3},\ c = 3$ $\\$ By using quadratic formula, we obtain $\\$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $\Rightarrow x = \dfrac{-4\sqrt{3} \pm \sqrt{48 - 48}}{8}$ $\\$ $\Rightarrow x = \dfrac{-4\sqrt{3} \pm 0}{8}$ $\\$ $\therefore x = \dfrac{-\sqrt{3}}{2}$ or $\dfrac{-\sqrt{3}}{2}$ $\\$

11   Find the roots of the following equations: $\\$ (i) $x - \dfrac{1}{x} = 3,\ x \ne 0$ $\\$ (ii) $\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30},\ x\ \ne\ – 4,\ 7$

##### Solution :

(ii) $\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$ $\\$ $\Rightarrow \dfrac{x - 7 - x - 4}{(x + 4)(x - 7)} = \dfrac{11}{30}$ $\\$ $\Rightarrow \dfrac{-11}{(x + 4)(x - 7)} = \dfrac{11}{30}$ $\\$ $\Rightarrow (x + 4)(x - 7) = - 30$ $\\$ $\Rightarrow x^2 - 3x - 28 = - 30$ $\\$ $\Rightarrow x^2 - 3x + 2 = 0$ $\\$ $\Rightarrow x^2 - 2x - x + 2 = 0$ $\\$ $\Rightarrow x(x - 2) - 1(x - 2) = 0$ $\\$ $\Rightarrow (x -2)(x - 1) = 0$ $\\$ $\Rightarrow x = 1$ or $2$

(i) $x - \dfrac{1}{x} = 3 \Rightarrow x^2 - 3x - 1 = 0$ $\\$ On comparing this equation with $ax^2 + bx + c = 0,$ we obtain $\\$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $\Rightarrow x = \dfrac{3 \pm \sqrt{9 + 4}}{2}$ $\\$ $\Rightarrow x = \dfrac{3 \pm \sqrt{13}}{2}$ $\\$ Therefore, $x = \dfrac{3 + \sqrt{13}}{2}$ or $\dfrac{3 - \sqrt{13}}{2}$ $\\$

12   The sum of the reciprocals of Rehman’s ages, (in years) $3$ years ago and $5$ years from now is $\dfrac{1}{3}$, Find his present age.

##### Solution :

Let the present age of Rehman be $x$ years. $\\$ Three years ago, his age was $(x - 3)$ years. $\\$ Five years hence, his age will be $(x + 5)$ years. $\\$ It is given that the sum of the reciprocals of Rehman’s ages $3$ years ago and $5$ years from now is $\dfrac{1}{3}$. $\\$ $\therefore \dfrac{1}{x - 3} + \dfrac{1}{x + 5} = \dfrac{1}{3}$ $\\$ $\dfrac{x + 5 + x - 3}{(x - 3)(x + 5)} = \dfrac{1}{3}$ $\\$ $\dfrac{2x + 2 }{(x - 3)(x + 5)} = \dfrac{1}{3}$ $\\$ $\Rightarrow 3(2X + 2) = (X - 3)(X + 5)$ $\\$ $\Rightarrow 6x + 6 = x^2 + 2x - 15$ $\\$ $\Rightarrow x^2 - 4x - 21$ $\\$ $\Rightarrow x^2 - 7x + 3x - 21 = 0$ $\\$ $\Rightarrow x(x - 7) + 3(x - 7) = 0$ $\\$ $\Rightarrow x = 7, - 3$ $\\$ However, age cannot be negative. $\\$ Therefore, Rehman’s present age is $7$ years.

13   In a class test, the sum of Shefali’s marks in Mathematics and English is $30$. Had she got $2$ marks more in Mathematics and $3$ marks less in English, the product of their marks would have been $210$. Find her marks in the two subjects.

##### Solution :

If the marks in Maths are $12$, then marks in English will be $30 - 12 = 18$ $\\$ If the marks in Maths are $13$, then marks in English will be $30 - 13 = 17$

Let the marks in Maths be $x.$ $\\$ Then, the marks in English will be $30 - x$. $\\$ According to the given question, $\\$ $(x + 2)(30 - x - 3) = 210$ $\\$ $(x + 2)(27 - x) = (210)$ $\\$ $\Rightarrow -x^2 + 25x + 54 = 210$ $\\$ $\Rightarrow x^2 - 25x + 156 = 0$ $\\$ $\Rightarrow x^2 - 25x - 13x + 156 = 0$ $\\$ $\Rightarrow x(x - 12) - 13(x - 12) = 0$ $\\$ $\Rightarrow (x - 12)(x - 13) = 0$ $\\$ $\Rightarrow x = 12,\ 13$

14   The diagonal of a rectangular field is $60\ metres$ more than the shorter side. If the longer side is $30\ metres$ more than the shorter side, find the sides of the field.

##### Solution :

$\Rightarrow x^2 + (x + 30)2^ = (x + 60)^2$ $\\$ $\Rightarrow x^2 + x^2 + 900 + 60x = x^2 + 3600 + 120x$ $\\$ $\Rightarrow x^2 - 60x - 2700 = 0$ $\\$ $\Rightarrow x^2 - 90x + 30x - 2700 = 0$ $\\$ $\Rightarrow x(x - 90) + 30(x - 90)$ $\\$ $\Rightarrow (x - 90)(x + 30) = 0$ $\\$ $\Rightarrow x = 90, -30$ $\\$ However, side cannot be negative. Therefore, the length of the shorter side will be $90\ m.$ $\\$ Hence, length of the larger side will be $(90 + 30)\ m = 120\ m$

Let the shorter side of the rectangle be $x\ m$. Then, larger side of the rectangle $= (x + 30)\ m$ $\\$ Diagonal of the rectangle $= \sqrt{x^2 + (x + 30)^2}$ $\\$ It is given that the diagonal of the rectangle is $60\ m$ more than the shorter side. $\\$ $\therefore \sqrt{x^2 + (x + 30)^2} = x + 60$ $\\$

15   The difference of squares of two numbers is $180$. The square of the smaller number is $8$ times the larger number. Find the two numbers.

##### Solution :

Let the larger and smaller number be $x$ and $y$ respectively. $\\$ According to the given question, $\\$ $x^2 - y^2 = 180$ and $y^2 = 8x$ $\\$ $\Rightarrow x^2 - 8x = 180$ $\\$ $\Rightarrow x^2 - 8x - 180 = 0$ $\\$ $\Rightarrow x^2 - 18x + 10x - 180 = 0$ $\\$ $\Rightarrow x(x - 18) - 10(x - 18) = 0$ $\\$ $\Rightarrow (x - 18)(x + 10) = 0$ $\\$ $\Rightarrow x = 18, -10$ $\\$ However, the larger number cannot be negative as $8$ times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible. Therefore, the larger number will be $18$ only. $x = 18$ $\\$ $\therefore y^2 = 8x = 8 \times 18 = 144$ $\\$ $\Rightarrow y = \pm \sqrt{144} = \pm 12$ $\\$ $\therefore Smaller\ number = \pm 12$ $\\$ Therefore, the numbers are $18$ and $12$ or $18$ and $-12$.

16   A train travels $360\ km$ at a uniform speed. If the speed had been $5\ km/h$ more, it would have taken $1\ hour$ less for the same journey. Find the speed of the train.

##### Solution :

Let the speed of the train be $x\ km/hr.$ $\\$ Time taken to cover $360\ km = \dfrac{360}{x}hr$ $\\$ According to the given question, $\\$ $(x + 5) \Big(\dfrac{360}{x} - 1\Big) = 360$ $\\$ $\Rightarrow (x + 5) \Big(\dfrac{360}{x} - 1\Big) = 360$ $\\$ $\Rightarrow 360 - x + \dfrac{1800}{x} - 5 = 360$ $\\$ $\Rightarrow x^2 + 5x - 1800 = 0$ $\\$ $\Rightarrow x^2 + 45x - 40x - 1800 = 0$ $\\$ $\Rightarrow (x + 45) - 40(x + 45) = 0$ $\\$ $\Rightarrow (x + 45)(x - 40) = 0$ $\\$ $\Rightarrow x = 40, - 45$ $\\$ However, speed cannot be negative. Therefore, the speed of train is $40\ km/h$

17   Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of larger diameter takes $10$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

##### Solution :

Time taken by the smaller pipe cannot be $\dfrac{30}{8} = 3.75$ hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible. $\\$ Therefore, time taken individually by the smaller pipe and the larger pipe will be $25$ and $25 - 10 = 15$ hours respectively.

Let the time taken by the smaller pipe to fill the tank be $x\ hr.$ $\\$ Time taken by the larger pipe $= (x - 10)\ hr$ $\\$ Part of tank filled by smaller pipe in $1$ hour = $\\$ Part of tank filled by larger pipe in $1$ hour = $\\$ It is given that the tank can be filled in $9\dfrac{3}{8} = \dfrac{75}{8}$ hours by both the pipes together. Therefore, $\\$ $\dfrac{1}{x} + \dfrac{1}{x - 10} = \dfrac{8}{75}$ $\\$ $\dfrac{x - 10 + x}{x(x - 10)} = \dfrac{8}{75}$ $\\$ $\Rightarrow \dfrac{2x - 10}{x(x - 10)} = \dfrac{8}{75}$ $\\$ $\Rightarrow 75(2x - 10) = 8x^2 - 80x$ $\\$ $\Rightarrow 150x - 750 = 8x^2 - 80x$ $\\$ $\Rightarrow 8x^2 - 230x + 750 = 0$ $\\$ $\Rightarrow 8x^2 - 200x - 30x + 750 = 0$ $\\$ $\Rightarrow 8x(x - 25) - 30(x - 25) = 0$ $\\$ $\Rightarrow(x - 25)(8x - 30) = 0$ $\\$ $i.e., x = 25, \dfrac{30}{8}$ $\\$

18   An express train takes $1\ hour$ less than a passenger train to travel $132\ km$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is $11\ km/h$ more than that of the passenger train, find the average speed of the two trains.

##### Solution :

Let the average speed of passenger train be $x\ km/h.$ $\\$ Average speed of express train $= (x + 11)\ km/h$ $\\$ It is given that the time taken by the express train to cover $132\ km$ is $1$ hour less than the passenger train to cover the same distance. $\\$ $\therefore \dfrac{132}{x} - \dfrac{132}{x + 11} = 1$ $\\$ $\Rightarrow 132 \Big[\dfrac{x + 11 - x}{x(x + 11)}\Big]$ $\\$ $\Rightarrow \dfrac{132 \times 11}{x(x + 11)} = 1$ $\\$ $\Rightarrow 132 \times 11 = x(x + 11)$ $\\$ $\Rightarrow x^2 + 11x - 1452 = 0$ $\\$ $\Rightarrow x^2 + 44x - 33x - 1452 = 0$ $\\$ $\Rightarrow x(x + 44) - 33(x + 44) = 0$ $\\$ $\Rightarrow (x + 44) ( x- 33)$ $\\$ $\Rightarrow x = -44, 33$ $\\$ Speed cannot be negative. $\\$ Therefore, the speed of the passenger train will be $33\ km/h$ and thus, the speed of the express train will be $33 + 11 = 44\ km/h$.

19   Sum of the areas of two squares is $468\ m^2.$ If the difference of their perimeters is $24\ m,$ find the sides of the two squares

##### Solution :

Also, $x^2 + y^2 = 468$ $\\$ $\Rightarrow (6 + y)^2 + y^2 = 468$ $\\$ $\Rightarrow 36 + y^2 + 12y + y^2 = 468$ $\\$ $\Rightarrow 2y^2 + 12y - 432 = 0$ $\\$ $\Rightarrow y^2 + 6y - 216$ $\\$ $\Rightarrow y^2 + 18y - 12y - 216 = 0$ $\\$ $\Rightarrow y(y + 18) - 12(y + 18)$ $\\$ $\Rightarrow (y + 18)(y -12)$ $\\$ $\Rightarrow y = -18$ or$12$ $\\$ However, side of a square cannot be negative. $\\$ Hence, the sides of the squares are $12\ m$ and $(12 + 6)\ m = 18\ m$

Let the sides of the two squares be $x\ m$ and $y\ m$. Therefore, their perimeter will be $4x$ and $4y$ respectively and their areas will be $x^2$ and $y^2$ respectively. $\\$ It is given that $\\$ $4x - 4y = 24$ $\\$ $x - y = 6$ $\\$ $x = y + 6$ $\\$

20   Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: $\\$ (i) $2x^2 – 3x + 5 = 0$ $\\$ (ii) $3x^2 – 4\sqrt{3}x + 4 = 0$ $\\$ (iii) $2x^2 - 6x + 3 = 0$

##### Solution :

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\\$ $= \dfrac{-(-6) \pm \sqrt{(-6)^2 -4(2)(3)}}{2(2)}$ $\\$ $= \dfrac{6 \pm \sqrt{12}}{4} = \dfrac{6 \pm 2\sqrt{3}}{4}$ $\\$ $= \dfrac{3 \pm \sqrt{3}}{2}$ $\\$ Therefore, the roots are $\dfrac{3 + \sqrt{3}}{2}$ or $\dfrac{3 - \sqrt{3}}{2}$.

We know that for a quadratic equation $ax^2 + bx + c = 0$, discriminant is $b^2 - 4ac.$ $\\$ (A) If $b^2 - 4ac > 0 \rightarrow$ two distinct real roots $\\$ (B) If $b^2 - 4ac = 0 \rightarrow$ two equal real roots $\\$ (C) If $b^2 - 4ac < 0 \rightarrow$ no real roots $\\$ (I) $2x^2 - 3x + 5 = 0$ $\\$ Comparing this equation with $ax^2 + bx + c = 0$, we obtain $a = 2,\ b = -3,\ c = 5$ $\\$ Discriminant $= b^2 - 4ac = (- 3)^2 - 4 (2) (5) = 9 - 40 = -31$ As $b^2 - 4ac < 0,$ $\\$ Therefore, no real root is possible for the given equation. $\\$ (II) $3x^2 - 4\sqrt{3x} + 4 = 0$ $\\$ Comparing this equation with $ax^2 + bx + c = 0$, we obtain $\\$ $a = 3,\ b = -4\sqrt{3},\ c = 4$ $\\$ Discriminant $= b^2 - 4ac = (-4\sqrt{3})^2 -4(3)(4)$ $\\$ $= 48 - 48 = 0$ $\\$ As $b^2 - 4ac = 0$, $\\$ Therefore, real roots exist for the given equation and they are equal to each other. $\\$ And the roots will be $\dfrac{-b}{2a}$ and $\dfrac{-b}{2a}$. $\\$ (III) $2x^2 - 6x + 3 = 0$ $\\$ Comparing this equation with $ax^2 + bx + c = 0,$ we obtain $a = 2,\ b = -6,\ c = 3$ $\\$ Discriminant $= b^2 - 4ac = (- 6)2 - 4 (2) (3)$ $\\$ $= 36 - 24 = 12$ $\\$ As $b^2 - 4ac > 0,$ $\\$ Therefore, distinct real roots exist for this equation as follows. $\\$

21   Find the values of k for each of the following quadratic equations, so that they have two equal roots. $\\$ (i) $2x^2 + kx + 3 = 0$ $\\$ (ii) $kx (x – 2) + 6 = 0$

##### Solution :

We know that if an equation $ax^2 + bx + c = 0$ has two equal roots, its discriminant $\\$ $(b^2 - 4ac)$ will be $0$. $\\$ (I) $2x^2 + kx + 3 = 0$ $\\$ Comparing equation with $ax^2 + bx + c = 0$, we obtain $a = 2,\ b = k,\ c = 3$ $\\$ Discriminant $= b^2 - 4ac = (k)^2 - 4(2) (3)$ $\\$ $= k^2 - 24$ For equal roots, $\\$ Discriminant $= 0$ $\\$ $k^2 - 24 = 0$ $\\$ $k^2 = 24$ $\\$ $k = \pm \sqrt{24} = \pm 2\sqrt{6}$ $\\$ (II) $kx (x - 2) + 6 = 0$ $\\$ or $kx^2 - 2kx + 6 = 0$ $\\$ Comparing this equation with $ax^2 + bx + c = 0$, we obtain $a = k,\ b = -2k,\ c = 6$ $\\$ Discriminant $= b^2 - 4ac = (- 2k)2 - 4 (k) (6)$ $\\$ $= 4k^2 - 24k$ For equal roots, $\\$ $b^2 - 4ac = 0$ $\\$ $4k^2 - 24k = 0$ $\\$ $4k\ (k - 6) = 0$ $\\$ Either $4k = 0$ or $k = 6 = 0$ $\\$ $k = 0$ or $k = 6$ $\\$ However, if $k = 0$, then the equation will not have the terms '$x^2$' and '$x$’. $\\$ Therefore, if this equation has two equal roots, $k$ should be $6$ only.

22   Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\ m^2$ ? If so, find its length and breadth.

##### Solution :

$l = \pm 20$ $\\$ However, length cannot be negative. $\\$ Therefore, breadth of mango grove $= 20\ m$ $\\$ Length of mango grove $= 2 \times 20 = 40\ m$ $\\$

Let the breadth of mango grove be $l$. $\\$ Area of mango grove $= (2l) (l)$ $\\$ $= 2l^2$ $\\$ $2l^2 = 800$ $\\$ $l^2 = \dfrac{800}{2} = 400$ $\\$ $l^2 - 400 = 0$ $\\$ Comparing this equation with $al^2 + bl + c = 0$, we obtain $a = 1,\ b = 0,\ c = 400$ $\\$ Discriminant $= b2 - 4ac = (0)2 - 4 \times (1) \times (- 400) = 1600$ $\\$ Here, $b^2 - 4ac > 0$ $\\$ Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed. $\\$

23   Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20$ years. Four years ago, the product of their ages in years was $48.$

##### Solution :

Let the age of one friend be $x$ years. $\\$ Age of the other friend will be $(20 - x)$ years. $\\$ $4$ years ago, age of $1^{st}$ friend $= (x - 4)$ years $\\$ And, age of $2^{nd}$ friend $= (20 - x - 4)$ $\\$ $= (16 - x)$ years $\\$ Given that, $\\$ $(x - 4) (16 - x) = 48$ $\\$ $16x - 64 - x2 + 4x = 48$ $\\$ $- x^2 + 20x - 112 = 0$ $\\$ $x^2 - 20x + 112 = 0$ $\\$ Comparing this equation with $ax^2 + bx + c = 0$, we obtain $\\$ $a = 1,\ b = -20,\ c = 112$ $\\$ $= 400 - 448 = -48$ $\\$ As $b^2 - 4ac < 0,$ $\\$ Therefore, no real root is possible for this equation and hence, this situation is not possible.

24   Is it possible to design a rectangular park of perimeter $80\ m$ and area $400\ m^2$ ? If so, find its length and breadth.

##### Solution :

Let the length and breadth of the park be $l$ and $b$. $\\$ Perimeter $= 2\ (l + b) = 80$ $l + b = 40$ $\\$ Or, $b = 40 - l$ $\\$ Area $= l \times b = l (40 - l) = 40l - l^2$ $\\$ $40l - l^2 = 400$ $\\$ $l^2 - 40l + 400 = 0$ $\\$ Comparing this equation with $\\$ $al^2 + bl + c = 0$, we obtain $\\$ $a = 1,\ b = -40,\ c = 400$ $\\$ Discriminate $= b^2 - 4ac = (- 40)2 -4 (1) (400)$ $\\$ $= 1600 - 1600 = 0$ $\\$ As $b^2 - 4ac = 0$,$\\$ Therefore, this equation has equal real roots. And hence, this situation is possible. $\\$ Root of this equation, $\\$ $l = -\dfrac{b}{2a}$ $\\$ $l = -\dfrac{(-40)}{2(1)} = \dfrac{40}{2} = 20$ $\\$ Therefore, length of park, $l = 20\ m$ $\\$ And breadth of park, $b = 40 - l = 40 - 20 = 20\ m$