**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

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### Question & Answer

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.3: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.4: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.5:Find two numbers whose sum is 27 and product is 182.

Ans :\(\begin{array}{l}{\text { Let the first number be } x \text { and the second number is } 27-x \text { . }} \\ {\text { Therefore, their product }=x(27-x)} \\ {\text { It is given that the product of these numbers is } 182 \text { . }} \\ {\text { Therefore, } x(27-x)=182}\end{array}\) \(\begin{array}{l}{\Rightarrow x^{2}-27 x+182=0} \\ {\Rightarrow x^{2}-13 x-14 x+182=0} \\ {\Rightarrow x(x-13)-14(x-13)=0} \\ {\Rightarrow(x-13)(x-14)=0}\end{array}\) \(\begin{array}{l}{\text { Either } x-13=0 \text { or } x-14=0} \\ {\text { l.e., } x=13 \text { or } x=14} \\ {\text { If first number }=13 \text { , then }} \\ {\text { other number }=27-13=14} \\ {\text { If first number }=14, \text { then }} \\ {\text { If first number }=27-14=13} \\ {\text { Therefore, the numbers are } 13 \text { and } 14 .}\end{array}\)

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