NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations – Here are all the NCERT solutions for Class 10 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete Chapter 4 titled Quadratic Equations of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 4 Quadratic Equations. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations in one place.

## NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations

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 Class 10 Subject Maths Book Mathematics Chapter Number 4 Chapter Name Quadratic Equations

### NCERT Solutions Class 10 Maths chapter 4 Quadratic Equations

Class 10, Maths chapter 4, Quadratic Equations solutions are given below in PDF format. You can view them online or download PDF file for future use.

Q.1: Check whether the following are quadratic equations :
$$(\text { i) } \quad(x+1)^{2}=2(x-3) \quad \text { (ii) } x^{2}-2 x=(-2)(3-x)$$
$$(\text { iii) } \quad(x-2)(x+1)=(x-1)(x+3) \quad \text { (iv) }(x-3)(2 x+1)=x(x+5)$$
$${ (v) }(2 x-1)(x-3)=(x+5)(x-1) \quad \text { (vi) } x^{2}+3 x+1=(x-2)^{2}$$
$${ (vii) }(x+2)^{3}=2 x\left(x^{2}-1\right) \quad \text { (viii) } x^{3}-4 x^{2}-x+1=(x-2)^{3}$$
Ans : $$\begin{array}{l}{\text { (i) } \quad(x+1)^{2}=2(x-3) \Rightarrow x^{2}+2 x+1=2 x-6 \Rightarrow x^{2}+7=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (ii) } \quad x^{2}-2 x=(-2)(3-x) \Rightarrow x^{2}-2 x=-6+2 x \Rightarrow x^{2}-4 x+6=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (iii) }(x-2)(x+1)=(x-1)(x+3) \Rightarrow x^{2}-x-2=x^{2}+2 x-3 \Rightarrow 3 x-1=0} \\ {\text { It is not of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is not a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (iv) } \quad(x-3)(2 x+1)=x(x+5) \Rightarrow 2 x^{2}-5 x-3=x^{2}+5 x \Rightarrow x^{2}-10 x-3=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (v) } \quad(2 x-1)(x-3)=(x+5)(x-1) \Rightarrow 2 x^{2}-7 x+3=x^{2}+4 x-5 \Rightarrow x^{2}-11 x+8=0} \\ {\text { of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (vi) } \quad x^{2}+3 x+1=(x-2)^{2} \Rightarrow x^{2}+3 x+1=x^{2}+4-4 x \Rightarrow 7 x-3=0} \\ {\text { It is not of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is not a quadratic equation. }} \\ {\text { (vii) } \quad(x+2)^{3}=2 x\left(x^{2}-1\right) \Rightarrow x^{3}+8+6 x^{2}+12 x=2 x^{3}-2 x \Rightarrow x^{3}-14 x-6 x^{2}-8=0} \\ {\text { not of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is not a quadratic equation. }}\end{array}$$
$$\begin{array}{l}{\text { (viii) } x^{3}-4 x^{3}-x+1=(x-2)^{3} \Rightarrow x^{3}-4 x^{2}-x+1=x^{3}-8-6 x^{2}+12 x \Rightarrow 2 x^{2}-13 x+9=0} \\ {\text { It is of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$$ 
Q.2: Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 $$\mathrm{m}^{2}$$ . 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.
Ans : $$\begin{array}{l}{\text { (i) Let the breadth of the plot be } x \mathrm{m} \text { . }} \\ {\text { Hence, the length of the plot is }(2 x+1) \mathrm{m} \text { . }} \\ {\text { Area of a rectangle = Length } \times \text { Breadth }} \\ {\therefore 528=x(2 x+1)} \\ {\Rightarrow 2 x^{2}+x-528=0}\end{array}$$
$$\begin{array}{l}{\text { (ii) Let the consecutive integers be } x \text { and } x+1} \\ {\text { It is given that their product is } 306 \text { . }} \\ {\therefore x(x+1)=306 \Rightarrow x^{2}+x-306=0}\end{array}$$
$$\begin{array}{l}{\text { (iii) Let Rohan's age be } x \text { . }} \\ {\text { Hence, his mother's age }=x+26} \\ {3 \text { years hence, }} \\ {\text { Rohan's age }=x+3} \\ {\text { Mother's age }=x+26+3=x+29} \\ {\text { It is given that the product of their ages after } 3 \text { years is } 360 \text { . }} \\ {\therefore(x+3)(x+29)=360} \\ {\Rightarrow x^{2}+32 x-273=0}\end{array}$$
$$\begin{array}{l}{\text { (iv) Let the speed of train be } x \mathrm{km} / \mathrm{h} \text { . }} \\ {\text { Time taken to travel } 480 \mathrm{km}=\frac{480}{x} \text { hrs }} \\ {\text { In second condition, let the speed of train }=(x-8) \mathrm{km} / \mathrm{h}} \\ {\text { It is also given that the train will take } 3 \text { hours to cover the same }} \\ {\text { distance. }} \\ {\text { Therefore, time taken to travel } 480 \mathrm{km}=\left(\frac{480}{x}+3\right)_{\text { hrs }}}\end{array}$$
$$\begin{array}{l}{\text { Speed } \times \text { Time }=\text { Distance }} \\ {(x-8)\left(\frac{480}{x}+3\right)=480} \\ {\Rightarrow 480+3 x-\frac{3840}{x}-24=480} \\ {\Rightarrow 3 x-\frac{3840}{x}=24} \\ {\Rightarrow 3 x^{2}-24 x+3840=0} \\ {\Rightarrow x^{2}-8 x+1280=0}\end{array}$$ 
Q.1: Find the roots of the following quadratic equations by factorisation:
$$\begin{array}{ll}{\text { (i) } x^{2}-3 x-10=0} & {\text { (ii) } 2 x^{2}+x-6=0} \\ {\text { (iii) }} & {\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0} \\ {\text { (v) } 100 x^{2}-20 x+1=0} & {\text { (iv) } 2 x^{2}-x+\frac{1}{8}=0}\end{array}$$
Ans : \begin{aligned}(i) & x^{2}-3 x-10 \\ &=x^{2}-5 x+2 x-10 \\ &=x(x-5)+2(x-5) \\ &=(x-5)(x+2) \end{aligned}
$$\begin{array}{l}{\text { Roots of this equation are the values for which }(x-5)(x+2)=0} \\ {\therefore x-5=0 \text { or } x+2=0} \\ {\text { i.e., } x=5 \text { or } x=-2}\end{array}$$

\begin{aligned}(\text { ii) }& 2 x^{2}+x-6 \\ &=2 x^{2}+4 x-3 x-6 \\ &=2 x(x+2)-3(x+2) \\ &=(x+2)(2 x-3) \end{aligned}
$$\begin{array}{l}{\text { Roots of this equation are the values for which }(x+2)(2 x-3)=0} \\ {\therefore x+2=0 \text { or } 2 x-3=0} \\ {\text { i.e., } x=-2 \text { or } x=\frac{3}{2}}\end{array}$$

\begin{aligned} \text { (iii) } & \sqrt{2} x^{2}+7 x+5 \sqrt{2} \\ &=\sqrt{2} x^{2}+5 x+5 x+5 \sqrt{2} \\ &=x(\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5) \\ &=(\sqrt{2} x+5)(x+\sqrt{2}) \end{aligned}
$$\begin{array}{l}{\text { Roots of this equation are the values for which }(\sqrt{2} x+5)(x+\sqrt{2})=0} \\ {\therefore \sqrt{2} x+5=0 \text { or } x+\sqrt{2}=0} \\ {\text { i.e., } x=\frac{-5}{\sqrt{2}} \text { or } x=-\sqrt{2}}\end{array}$$

(iv)
\begin{aligned} & 2 x^{2}-x+\frac{1}{8} \\ &=\frac{1}{8}\left(16 x^{2}-8 x+1\right) \\ &=\frac{1}{8}\left(16 x^{2}-4 x-4 x+1\right) \\ &=\frac{1}{8}(4 x(4 x-1)-1(4 x-1)) \\ &=\frac{1}{8}(4 x-1)^{2} \end{aligned}
$$\begin{array}{l}{\text { Roots of this equation are the values for which }(4 x-1)^{2}=0} \\ {\text { Therefore, }(4 x-1)=0 \text { or }(4 x-1)=0} \\ {\text { i.e., } x=\frac{1}{4} \text { or } x=\frac{1}{4}}\end{array}$$

\begin{aligned}(v) & 100 x^{2}-20 x+1 \\ &=100 x^{2}-10 x-10 x+1 \\ &=10 x(10 x-1)-1(10 x-1) \\ &=(10 x-1)^{2} \end{aligned}
$$\begin{array}{l}{\text { Roots of this equation are the values for which }(10 x-1)^{2}=0} \\ {\text { Therefore, }(10 x-1)=0 \text { or }(10 x-1)=0} \\ {\text { i.e., } x=\frac{1}{10} \text { or } x=\frac{1}{10}}\end{array}$$ 
Q.2: Solve the problems:
Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
Ans : $$\begin{array}{l}{\text { (i) Let the number of John's marbles be } x \text { . }} \\ {\text { The refore, number of Jivanti's marble }=45-x} \\ {\text { After losing } 5 \text { marbles, }} \\ {\text { Number of John's marbles = } x-5} \\ {\text { Number of Jivanti's marbles = } 45-x-5=40-x} \\ {\text { It is given that the product of their marbles is } 124 .}\end{array}$$
$$\begin{array}{l}{\therefore(x-5)(40-x)=124} \\ {\Rightarrow x^{2}-45 x+324=0} \\ {\Rightarrow x^{2}-36 x-9 x+324=0} \\ {\Rightarrow x(x-36)-9(x-36)=0} \\ {\Rightarrow(x-36)(x-9)=0}\end{array}$$
$$\begin{array}{l}{\text { Either } x-36=0 \text { or } x-9=0} \\ {\text { i.e., } x=36 \text { or } x=9} \\ {\text { If the number of John's marbles }=36 \text { , }} \\ {\text { Then, number of Jivanti's marbles }=45-36=9} \\ {\text { If number of John's marbles }=9,} \\ {\text { Then, number of Jivanti's marbles }=45-9=36}\end{array}$$

$$\begin{array}{l}{\text { (ii) Let the number of toys produced be } x \text { . }} \\ {\therefore \text { cost of production of each toy }=R s(55-x)} \\ {\text { It is given that, total production of the toys }=Rs 750}\end{array}$$
$$\begin{array}{l}{\therefore x(55-x)=750} \\ {\Rightarrow x^{2}-55 x+750=0} \\ {\Rightarrow x^{2}-25 x-30 x+750=0} \\ {\Rightarrow x(x-25)-30(x-25)=0} \\ {\Rightarrow(x-25)(x-30)=0}\end{array}$$
$$\begin{array}{l}{\text { Either } x-25=0 \text { or } x-30=0} \\ {\text { i.e., } x=25 \text { or } x=30} \\ {\text { Hence, the number of toys will be either } 25 \text { or } 30 \text { . }}\end{array}$$ 
Q.3: Find two numbers whose sum is 27 and product is 182.
Ans : $$\begin{array}{l}{\text { Let the first number be } x \text { and the second number is } 27-x \text { . }} \\ {\text { Therefore, their product }=x(27-x)} \\ {\text { It is given that the product of these numbers is } 182 \text { . }} \\ {\text { Therefore, } x(27-x)=182}\end{array}$$
$$\begin{array}{l}{\Rightarrow x^{2}-27 x+182=0} \\ {\Rightarrow x^{2}-13 x-14 x+182=0} \\ {\Rightarrow x(x-13)-14(x-13)=0} \\ {\Rightarrow(x-13)(x-14)=0}\end{array}$$
$$\begin{array}{l}{\text { Either } x-13=0 \text { or } x-14=0} \\ {\text { l.e., } x=13 \text { or } x=14} \\ {\text { If first number }=13 \text { , then }} \\ {\text { other number }=27-13=14} \\ {\text { If first number }=14, \text { then }} \\ {\text { If first number }=27-14=13} \\ {\text { Therefore, the numbers are } 13 \text { and } 14 .}\end{array}$$ 
Q.4: Find two consecutive positive integers, sum of whose squares is 365.
Ans : $$\begin{array}{l}{\text { Let the consecutive positive integers be } x \text { and } x+1 .} \\ {\text { Given that } x^{2}+(x+1)^{2}=365} \\ {\Rightarrow x^{2}+x^{2}+1+2 x=365} \\ {\Rightarrow 2 x^{2}+2 x-364=0} \\ {\Rightarrow x^{2}+x-182=0} \\ {\Rightarrow x^{2}+14 x-13 x-182=0} \\ {\Rightarrow x(x+14)(x-13)=0} \\ {\Rightarrow(x+14)(x-13)=0}\end{array}$$
$$\begin{array}{l}{\text { Either } x+14=0 \text { or } x-13=0, \text { l.e., } x=-14 \text { or } x=13} \\ {\text { since the integers are positive, } x \text { can only be } 13 .} \\ {\therefore x+1=13+1=14} \\ {\text { Therefore, two consecutive positive integers will be } 13 \text { and } 14 .}\end{array}$$ 
Q.5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Ans : \begin{array}{l}{\text { Let the base of the right triangle be } x \mathrm{cm} .} \\ {\text { Its altitude }=(x-7) \mathrm{cm}}\end{array}
$$\begin{array}{l}{\text { From pythagoras theorem, }} \\ {\text { Base }^{2}+\text { Altitude }^{2}=\text { Hypotenuse }^{2}} \\ {\therefore x^{2}+(x-7)^{2}=13^{2}} \\ {\Rightarrow x^{2}+x^{2}+49-14 x=169} \\ {\Rightarrow 2 x^{2}-14 x-120=0} \\ {\Rightarrow x^{2}-7 x-120=0} \\ {\Rightarrow x^{2}-7 x-120=0} \\ {\Rightarrow x^{2}-7 x-60=0} \\ {\Rightarrow x^{2}-12 x-5 x-60=0} \\ {\Rightarrow x(x-12)+5(x-12)=0}\end{array}$$
$$\begin{array}{l}{\Rightarrow(x-12)(x+5)=0} \\ {\text { Either } x-12=0 \text { or } x+5=0, \text { i.e., } x=12 \text { or } x=-5} \\ {\text { since sides are positive, } x \text { can only be } 12 \text { . }} \\ {\text { Therefore, the base of the given triangle is } 12 \mathrm{cm} \text { and the altitude of }} \\ {\text { this triangle will be }(12-7) \mathrm{cm}=5 \mathrm{cm} .}\end{array}$$ 
Q.6: 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.
Ans : $$\begin{array}{l}{\text { Let the number of articles produced be } x \text { . }} \\ {\text { Therefore, cost of production of each article = Rs }(2 x+3)} \\ {\text { It is given that the total production is Rs 90. }}\end{array}$$
$$\begin{array}{l}{\therefore x(2 x+3)=90} \\ {\Rightarrow 2 x^{2}+3 x-90=0} \\ {\Rightarrow 2 x^{2}+15 x-12 x-90=0} \\ {\Rightarrow x(2 x+15)-6(2 x+15)=0} \\ {\Rightarrow(2 x+15)(x-6)=0}\end{array}$$
$$\begin{array}{l}{\text { Either } 2 x+15=0 \text { or } x-6=0, \text { i.e., } x=\frac{-15}{2} \text { or } x=6} \\ {\text { As the number of articles produced can only be a positive integer, }} \\ {\text { therefore, } x \text { can only be } 6 \text { . }} \\ {\text { Hence, number of articles produced }=6} \\ {\text { cost of each article }=2 \times 6+3=\operatorname{Rs} 15}\end{array}$$ 
Q.1: Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
$$\begin{array}{ll}{\text { (i) } 2 x^{2}-7 x+3=0} & {\text { (iii) }} & {4 x^{2}+4 \sqrt{3} x+3=0}\end{array} \quad \begin{array}{ll}{\text { (ii) } 2 x^{2}+x-4=0} \\ {\text { (iv) } 2 x^{2}+x+4=0}\end{array}$$
Ans : (i)
$$\begin{array}{l}{2 x^{2}-7 x+3=0} \\ {\Rightarrow 2 x^{2}-7 x=-3} \\ {\text { On dividing both sides of the equation by } 2, \text { we obtain }} \\ {\Rightarrow x^{2}-\frac{7}{2} x=-\frac{3}{2}} \\ {\Rightarrow x^{2}-2 \times x \times \frac{7}{4}=-\frac{3}{2}} \\ {\text { On adding }\left(\frac{7}{4}\right)^{2} \text { to both sides of equation, we obtain }}\end{array}$$
$$\begin{array}{l}{\Rightarrow(x)^{2}-2 \times x \times \frac{7}{4}+\left(\frac{7}{4}\right)^{2}=\left(\frac{7}{4}\right)^{2}-\frac{3}{2}} \\ {\Rightarrow\left(x-\frac{7}{4}\right)^{2}=\frac{49}{16}-\frac{3}{2}} \\ {\Rightarrow\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}} \\ {\Rightarrow\left(x-\frac{7}{4}\right)=\pm \frac{5}{4}} \\ {\Rightarrow x=\frac{7}{4} \pm \frac{5}{4}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{7}{4}+\frac{5}{4} \text { or } x=\frac{7}{4}-\frac{5}{4}} \\ {\Rightarrow x=\frac{12}{4} \text { or } x=\frac{2}{4}} \\ {\Rightarrow x=3 \text { or } \frac{1}{2}}\end{array}$$
$$\begin{array}{l}{\text { (ii) } 2 x^{2}+x-4=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=1, c=-4} \\ {\text { By using quadratic formula, we obtain }}\end{array}$$
$$\begin{array}{l}{x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{1+32}}{4}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{33}}{4}} \\ {\therefore x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}}\end{array}$$
$$\begin{array}{l}{\text { (iii) } 4 x^{2}+4 \sqrt{3} x+3=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {q=4, b=4 \sqrt{3}, c=3} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}} \\ {\Rightarrow x=\frac{-4 \sqrt{3} \pm 0}{8}} \\ {\therefore x=\frac{-\sqrt{3}}{2} \text { or } \frac{-\sqrt{3}}{2}}\end{array}$$
$$\begin{array}{l}{\text { (iv) } 2 x^{2}+x+4=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=1, c=4} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{-1 \pm \sqrt{1-32}}{4}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{-31}}{4}} \\ {\text { However, the square of a number cannot be negative. }} \\ {\text { Therefore, there is no real root for the given equation. }}\end{array}$$ 
Q.2: Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Ans : $$\begin{array}{l}{\text { (i) } 2 x^{2}-7 x+3=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=-7 . c=3} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {\Rightarrow x=\frac{7 \pm \sqrt{49-24}}{4}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{7 \pm \sqrt{25}}{4}} \\ {\Rightarrow x=\frac{7 \pm 5}{4}} \\ {\Rightarrow x=\frac{7+5}{4} \text { or } \frac{7-5}{4}} \\ {\Rightarrow x=\frac{12}{4} \text { or } \frac{2}{4}} \\ {\therefore x=3 \text { or } \frac{1}{2}}\end{array}$$
$$\begin{array}{l}{\text { (ii) } 2 x^{2}+x-4=0} \\ {\text { On comparing thisequation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=1, c=-4} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{1+32}}{4}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{33}}{4}}\end{array}$$
$$\therefore x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}$$
$$\begin{array}{l}{\text { (iii) } 4 x^{2}+4 \sqrt{3} x+3=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=4, b=4 \sqrt{3}, c=3} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {\Rightarrow x=\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{-4 \sqrt{3} \pm 0}{8}} \\ {\therefore x=\frac{-\sqrt{3}}{2} \text { or } \frac{-\sqrt{3}}{2}}\end{array}$$
$$\begin{array}{l}{\text { (iv) } 2 x^{2}+x+4=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=1, c=4} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{-1 \pm \sqrt{1-32}}{4}} \\ {\Rightarrow x=\frac{-1 \pm \sqrt{-31}}{4}} \\ {\text { However, the square of a number cannot be negative. }} \\ {\text { Therefore, there is no real root for the given equation. }}\end{array}$$ 
Q.3: Find the roots of the following equations:
(i) $$\quad x-\frac{1}{x}=3, x \neq 0$$
(ii) $$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7$$
Ans : $$\begin{array}{l}{\text { (i) } \quad x-\frac{1}{x}=3 \Rightarrow x^{2}-3 x-1=0} \\ {\text { On comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=1, b=-3, c=-1} \\ {\text { By using quadratic formula, we obtain }} \\ {x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}$$
$$\begin{array}{l}{\Rightarrow x=\frac{3 \pm \sqrt{9+4}}{2}} \\ {\Rightarrow x=\frac{3 \pm \sqrt{13}}{2}} \\ {\text { Therefore, } x=\frac{3+\sqrt{13}}{2} \text { or } \frac{3-\sqrt{13}}{2}}\end{array}$$
\begin{aligned} \text { (ii) } & \frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30} \\ & \Rightarrow \frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30} \\ & \Rightarrow \frac{-11}{(x+4)(x-7)}=\frac{11}{30} \\ & \Rightarrow(x+4)(x-7)=-30 \\ & \Rightarrow x^{2}-3 x-28=-30 \end{aligned}
$$\begin{array}{l}{\Rightarrow x^{2}-3 x+2=0} \\ {\Rightarrow x^{2}-2 x-x+2=0} \\ {\Rightarrow x(x-2)-1(x-2)=0} \\ {\Rightarrow(x-2)(x-1)=0} \\ {\Rightarrow x=1 \text { or } 2}\end{array}$$ 
Q.4: The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $$\frac{1}{3}$$ Find his present age.
Ans : $$\begin{array}{l}{\text { Let the present age of Rehman be } x \text { years. }} \\ {\text { Three years ago, his age was }(x-3) \text { years. }} \\ {\text { Five years hence, his age will be }(x+5) \text { years. }} \\ {\text { It is given that the sum of the reciprocals of Rehman's ages 3 years }} \\ {\text { ago and } 5 \text { years from now is } \frac{1}{3}}\end{array}$$
$$\begin{array}{l}{\therefore \frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}} \\ {\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}} \\ {\frac{2 x+2}{(x-3)(x+5)}=\frac{1}{3}} \\ {\Rightarrow 3(2 x+2)=(x-3)(x+5)}\end{array}$$
$$\begin{array}{l}{\Rightarrow 6 x+6=x^{2}+2 x-15} \\ {\Rightarrow x^{2}-4 x-21=0} \\ {\Rightarrow x^{2}-7 x+3 x-21=0} \\ {\Rightarrow x(x-7)+3(x-7)=0} \\ {\Rightarrow(x-7)(x+3)=0} \\ {\Rightarrow x=7,-3} \\ {\text { However, age cannot be negative. }} \\ {\text { Therefore, Rehman's present age is } 7 \text { years. }}\end{array}$$ 
Q.5: 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.
Ans : $$\begin{array}{l}{\text { Let the marks in Maths be } x \text { . }} \\ {\text { Then, the marks in English will be } 30-x \text { . }} \\ {\text { According to the given question, }} \\ {(x+2)(30-x-3)=210} \\ {(x+2)(27-x)=210}\end{array}$$
$$\begin{array}{l}{\Rightarrow-x^{2}+25 x+54=210} \\ {\Rightarrow x^{2}-25 x+156=0} \\ {\Rightarrow x^{2}-12 x-13 x+156=0} \\ {\Rightarrow x(x-12)-13(x-12)=0} \\ {\Rightarrow(x-12)(x-13)=0} \\ {\Rightarrow x=12,13}\end{array}$$
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 
Q.6: 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.
Ans : $$\begin{array}{l}{\text { Let the shorter side of the rectangle be } x \text { m. }} \\ {\text { Then, larger side of the rectangle = }(x+30) \mathrm{m}} \\ {\text { Diagonal of the rectangle }=\sqrt{x^{2}+(x+30)^{2}}} \\ {\text { It is given that the diagonal of the rectangle is } 60 \mathrm{m} \text { more than the shorter side. }}\end{array}$$
$$\begin{array}{l}{\therefore \sqrt{x^{2}+(x+30)^{2}}=x+60} \\ {\Rightarrow x^{2}+(x+30)^{2}=(x+60)^{2}} \\ {\Rightarrow x^{2}+x^{2}+900+60 x=x^{2}+3600+120 x} \\ {\Rightarrow x^{2}-60 x-2700=0} \\ {\Rightarrow x^{2}-90 x+30 x-2700=0} \\ {\Rightarrow(x-90)(x+30)=0} \\ {\Rightarrow x=90,-30}\end{array}$$
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 
Q.7: 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.
Ans : Let the larger and smaller number be x and y respectively. According to the given question,
$$\begin{array}{l}{x^{2}-y^{2}=180 \text { and } y^{2}=8 x} \\ {\Rightarrow x^{2}-8 x=180} \\ {\Rightarrow x^{2}-8 x-180=0} \\ {\Rightarrow x^{2}-18 x+10 x-180=0} \\ {\Rightarrow x(x-18)+10(x-18)=0} \\ {\Rightarrow(x-18)(x+10)=0} \\ {\Rightarrow x=18,-10}\end{array}$$
However,the larger number can not  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.
$$\begin{array}{l}{x=18} \\ {\therefore y^{2}=8 x=8 \times 18=144} \\ {\Rightarrow y=\pm \sqrt{144}=\pm 12} \\ {\therefore \text { Smaller number }=\pm 12} \\ {\text { Therefore, the numbers are } 18 \text { and } 12 \text { or } 18 \text { and }-12 \text { . }}\end{array}$$ 
Q.8: 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.
Ans : $$\begin{array}{l}{\text { Let the speed of the train be } x \mathrm{km} / \mathrm{hr} \text { . }} \\ {\text { Time taken to cover } 360 \mathrm{km}=\frac{360}{x \text { hr }}} \\ {\text { According to the given question, }}\end{array}$$
$$\begin{array}{l}{(x+5)\left(\frac{360}{x}-1\right)=360} \\ {\Rightarrow(x+5)\left(\frac{360}{x}-1\right)=360} \\ {\Rightarrow 360-x+\frac{1800}{x}-5=360} \\ {\Rightarrow x^{2}+5 x-1800=0} \\ {\Rightarrow x^{2}+45 x-1800=0} \\ {\Rightarrow x(x+45)(x-40)=0} \\ {\Rightarrow(x+45)(x-40)=0} \\ {\Rightarrow x=40,-45}\end{array}$$
However, speed cannot be negative.Therefore, the speed Of train is 40 km/h. 
Q.9: Two water taps together can fill a tank in $$9\frac{3}{8}$$ 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.
Ans : 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 = $$\frac{1}{x}$$
Part of tank filled by smaller pipe in 1 hour = $$\frac{1}{x-10}$$
$$\begin{array}{l}{\text { It is given that the tank can be filled in } 9 \frac{3}{8}=\frac{75}{8} \text { hours by both the pipes }} \\ {\text { together. Therefore, }} \\ {\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}} \\ {\frac{x-10+x}{x(x-10)}=\frac{8}{75}} \\ {\Rightarrow \frac{2 x-10}{x(x-10)}=\frac{8}{75}}\end{array}$$
$$\begin{array}{l}{\Rightarrow 75(2 x-10)=8 x^{2}-80 x} \\ {\Rightarrow 150 x-750=8 x^{2}-80 x} \\ {\Rightarrow 8 x^{2}-230 x+750=0} \\ {\Rightarrow 8 x^{2}-200 x-30 x+750=0} \\ {\Rightarrow 8 x(x-25)-30(x-25)=0} \\ {\Rightarrow(x-25)(8 x-30)=0} \\ {\text { i.e.. } x=25, \frac{30}{8}}\end{array}$$
Time taken by the smaller pipe cannot be 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. 
Q.10: 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 11km/h more than that of the passenger train, find the average speed of the two trains. 11. Sum of the areas of two squares is 468 m² . If the difference of their perimeters is 24 m, find the sides of the two squares.
Ans : 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.
$$\begin{array}{l}{\therefore \frac{132}{x}-\frac{132}{x+11}=1} \\ {\Rightarrow 132\left[\frac{x+11-x}{x(x+11)}\right]=1} \\ {\Rightarrow \frac{132 \times 11}{x(x+11)}=1} \\ {\Rightarrow 132 \times 11=x(x+11)} \\ {\Rightarrow x^{2}+11 x-1452=0}\end{array}$$
$$\begin{array}{l}{\Rightarrow x^{2}+44 x-33 x-1452=0} \\ {\Rightarrow x(x+44)-33(x+44)=0} \\ {\Rightarrow(x+44)(x-33)=0} \\ {\Rightarrow(x+44)(x-33)=0} \\ {\Rightarrow x=-44,33} \\ {\text { Speed cannot be negative. }} \\ {\text { Therefore, the speed of the passenger train will be } 33 \mathrm{km} / \mathrm{h} \text { and thus, }} \\ {\text { the speed of the express train will be } 33+11=44 \mathrm{km} / \mathrm{h} \text { . }}\end{array}$$ 
Q.11: Sum of the areas of two squares is 468  m² . If the difference of their perimeters is 24 m, find the sides of the two squares.
Ans : 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
$$\begin{array}{l}{4 x-4 y=24} \\ {x-y=6} \\ {x=y+6} \\ {\text { Also, } x^{2}+y^{2}=468} \\ {\Rightarrow(6+y)^{2}+y^{2}=468} \\ {\Rightarrow 36+y^{2}+12 y+y^{2}=468}\end{array}$$
$$\begin{array}{l}{\Rightarrow 2 y^{2}+12 y-432=0} \\ {\Rightarrow y^{2}+6 y-216=0} \\ {\Rightarrow y^{2}+18 y-12 y-216=0} \\ {\Rightarrow y(y+18)-12(y+18)=0} \\ {\Rightarrow(y+18)(y-12)=0} \\ {\Rightarrow y=-18 \text { or } 12}\end{array}$$
However, side Of a square cannot be negative, Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m. 
Q.1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
$$\begin{array}{l}{\text { (I) } 2 x^{2}-3 x+5=0} \\ {\text { (II) } 3 x^{2}-4 \sqrt{3} x+4=0} \\ {\text { (III) } 2 x^{2}-6 x+3=0}\end{array}$$
Ans : We know that for a quadratic equation $$a x^{2}+b x+c=0$$, discriminant is  b²  - 4ac.
$$\begin{array}{l}{\text { (A) If } b^{2}-4 a c>0 \rightarrow \text { two distinct real roots }} \\ {\text { (B) If } b^{2}-4 a c=0 \rightarrow \text { two equal real roots }} \\ {\text { (C) If } b^{2}-4 a c<0 \rightarrow \text { no real roots }} \\ {\text { (I) } 2 x^{2}-3 x+5=0}\end{array}$$
$$\begin{array}{l}{\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=-3, c=5} \\ {\text { Discriminant }=b^{2}-4 a c=(-3)^{2}-4(2)(5)=9-40} \\ {=-31} \\ {\text { As } b^{2}-4 a c<0} \\ {\text { Therefore, no real root is possible for the given equation. }} \\ {\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=3, b=-4 \sqrt{3}, c=4}\end{array}$$
$$\begin{array}{l}{\text { Discriminant }=b^{2}-4 a c=(-4 \sqrt{3})^{2}-4(3)(4)} \\ {=48-48=0} \\ {\text { As } b^{2}-4 a c=0} \\ {\text { Therefore, real roots exist for the given equation and they are equal to }} \\ {\text { each other. }} \\ {\text { And the roots will be } \frac{-b}{2 a} \text { and } \frac{-b}{2 a}}\end{array}$$
$$\begin{array}{l}{\frac{-b}{2 a}=\frac{-(-4 \sqrt{3})}{2 \times 3}=\frac{4 \sqrt{3}}{6}=\frac{2 \sqrt{3}}{3}=\frac{2}{\sqrt{3}}} \\ {\text { Therefore, the roots are } \sqrt{3} \text { and } \frac{2}{\sqrt{3}}}\end{array}$$
$$\begin{array}{l}{\text { (III) } 2 x^{2}-6 x+3=0} \\ {\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=-6, c=3} \\ {\text { Discriminant }=b^{2}-4 a c=(-6)^{2}-4(2)(3)} \\ {=36-24=12} \\ {\text { As } b^{2}-4 a c>0} \\ {\text { Therefore, distinct real roots exist for this equation as follows. }}\end{array}$$
$$\begin{array}{l}{x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\ {\quad=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(2)(3)}}{2(2)}} \\ {=\frac{6 \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4}} \\ {=\frac{3 \pm \sqrt{3}}{2}} \\ {\text { Therefore, the roots are } \frac{3+\sqrt{3}}{2} \text { or } \frac{3-\sqrt{3}}{2}}\end{array}$$ 
Q.2: Find the values of k for each of the following quadratic equations, so that they have two equal roots.
$$\begin{array}{l}{\text { (I) } 2 x^{2}+k x+3=0} \\ {\text { (II) } k x(x-2)+6=0}\end{array}$$
Ans : We know that if an equation $$a x^{2}+b x+c=0$$has two equal roots, its discriminant
$$\begin{array}{l}{\left(b^{2}-4 a c\right) \text { will be } 0 .} \\ {\text { (I) } 2 x^{2}+k x+3=0}\end{array}$$
$$\begin{array}{l}{\text { Comparing equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=2, b=k, c=3} \\ {\text { Discriminant }=b^{2}-4 a c=(k)^{2}-4(2)(3)} \\ {=k^{2}-24} \\ {\text { For equal roots, }}\end{array}$$
$$\begin{array}{l}{\text { Discriminant }=0} \\ {k^{2}-24=0} \\ {k^{2}=24} \\ {k=\pm \sqrt{24}=\pm 2 \sqrt{6}}\end{array}$$
$$\begin{array}{l}{\text { (II) } k x(x-2)+6=0} \\ {\text { or } k x^{2}-2 k x+6=0} \\ {\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=k, b=-2 k, c=6} \\ {\text { Discriminant }=b^{2}-4 a c=(-2 k)^{2}-4(k)(6)} \\ {=4 k^{2}-24 k}\end{array}$$
$$\begin{array}{l}{\text { For equal roots, }} \\ {b^{2}-4 a c=0} \\ {4 k^{2}-24 k=0} \\ {4 k(k-6)=0} \\ {\text { Either } 4 k=0 \text { or } k=6=0} \\ {k=0 \text { or } k=6}\end{array}$$
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. 
Q.3: 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.
Ans : Let the breadth of mango grove be l.Length of mango grove will be 21. Area of mango grove = (21) (l)
$$\begin{array}{l}{=2 l^{2}} \\ {2 l^{2}=800} \\ {l^{2}=\frac{800}{2}=400} \\ {l^{2}-400=0} \\ {\text { Comparing this equation with } a l^{2}+bl+c=0, \text { we obtain }} \\ {a=1 b=0, c=400} \\ {\text { Discriminant }=b^{2}-4 a c=(0)^{2}-4 \times(1) \times(-400)=1600} \\ {\text { Here, } b^{2}-4 a c>0}\end{array}$$
$$\begin{array}{l}{\text { Therefore, the equation will have real roots. And hence, the desired }} \\ {\text { rectangular mango grove can be designed. }} \\ {l=\pm 20} \\ {\text { However, length cannot be negative. }} \\ {\text { Therefore, breadth of mango grove = 20 } \mathrm{m}} \\ {\text { Length of mango grove }=2 \times 20=40 \mathrm{m}}\end{array}$$ 
Q.4: 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.
Ans : Let the age of one friend be x years. Age of the other friend will be (20 - x) years. 4 years ago, age of 1st friend = (x - 4) years
$$\begin{array}{l}{\text { And, age of } 2^{\text { nd }} \text { friend }=(20-x-4)} \\ {=(16-x) \text { years }} \\ {\text { Given that, }} \\ {(x-4)(16-x)=48} \\ {16 x-64-x^{2}+4 x=48} \\ {-x^{2}+20 x-112=0} \\ {x^{2}-20 x+112=0}\end{array}$$
$$\begin{array}{l}{\text { Comparing this equation with } a x^{2}+b x+c=0, \text { we obtain }} \\ {a=1, b=-20, c=112} \\ {\text { Discriminant }=b^{2}-4 a c=(-20)^{2}-4(1)(112)} \\ {=400-448=-48} \\ {\text { As } b^{2}-4 a c<0} \\ {\text { Therefore, no real root is possible for this equation and hence, this }} \\ {\text { situation is not possible. }}\end{array}$$ 
Q.5: Is it possible to design a rectangular park of perimeter 80 and area $$400 m^{2}$$? If so find its length and breadth.
Ans : Let the length and breadth of the park be I and b. Perimeter = 2 (l + b) = 80  l + b = 40  or, b = 40 - l
$$\begin{array}{l}{\text { Area }=l \times b=I(40-l)=40 l-l^{2}} \\ {40l-l^{2}=400} \\ {P-40 l+400=0}\end{array}$$
$$\begin{array}{l}{\text { Comparing this equation with }} \\ {a f+b l+c=0, \text { we obtain }} \\ {a=1, b=-40, c=400} \\ {\text { Discriminate }=b^{2}-4 a c=(-40)^{2}-4(1)(400)} \\ {=1600-1600=0} \\ {\text { As } b^{2}-4 a c=0}\end{array}$$
$$\begin{array}{l}{\text { Therefore, this equation has equal real roots. And hence, this situation }} \\ {\text { is possible. }} \\ {\text { Root of this equation, }} \\ {l=-\frac{b}{2 a}} \\ {l=-\frac{(-40)}{2(1)}=\frac{40}{2}=20} \\ {\text { Therefore, length of park, } I=20 \mathrm{m}} \\ {\text { And breadth of park, } b=40-I=40-20=20 \mathrm{m}}\end{array}$$ 

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