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Quadratic Equations is an important chapter in the Class Xth Maths syllabus, which lays the foundation for solving complex mathematical problems. This chapter deals with the concept of quadratic equations and their various applications in real-life situations. In this article, we will discuss the NCERT Solution for Class Xth Maths Chapter 4 Quadratic Equations by Swastik Classes. These solutions provide comprehensive explanations and step-by-step solutions to all the problems in the textbook, making it easier for students to understand the concepts and excel in their exams. In the following sections, we will provide an overview of the topics covered in this chapter and how the NCERT solutions by Swastik Classes can help students in their preparation.

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Answers of Maths NCERT solutions for Class Xth Maths Chapter 4 Quadratic Equations

 Unit 4

Quadratic Equations 

Exercise 4.1 

  Q1 :

Check whether the following are quadratic equations:c8UaueUoj9I2Nif8wUJIMHMIHV3lpPMUsqKlf1jAUqqusovDh1CCFr4LxkN6viGnYvKrtyv62imOhgjIzOAjlvmvlvSKs XKBPzXbumSozj6CXHwrQ7DNLWDanO3X9KF6Nbc9tQ

6aa8legud HpwZGVIp OgONAii7DZJRQt3cnlKGiaRkVw0gjLBwdXmK VRqwIuphJtzslG0HfcoYlef5H7f5JekCo7DYy2VryA9TRYxA aseSY3okeAYCBYPuSe IW9HVjeJSnc

DjvJ1MAPgpGuNmcbfCxX3PW PujAOpTcI9RxvDOWnVO8gaSmUUN5rrPxtuO0Q03DI pcJMD MWl 6Jv6xPTzd495 S FpiXj3TBfOsrOImmKcg1AMldKvX9 c03Ch8Ol qG6Cfg

u5RjwLGZ8cH5GjxeoXQY58 I8WusvOvPdq3

Answer:XEmNpjbhQBcEz7cH4JlF20Kj5vRBGYXCDItyQvvklg63ych6zp kkxge2aHQZ7t0FfDWk0WBR4qhvFq 5JxeYxWzh9ylafgzpBgRYEFok3CNZGE31CtJM6R TT8 0Fsov4GUz2w

It is of the form

Hence, the given equation is a quadratic equation.83kmYWxWEKXEd3ngsl5q9JDf hFKBXwcxo5fPeyfyrMFL6TKZb5 36LL2T292ebSbIGaOySS40CH96YfFr351J786rP

It is of the form .

Hence, the given equation is a quadratic equation.8kSybEl3iNuXLM2aSm mNDcjg8t511x7OdOYIkrS a1vn42DjrkahGbl APEEVH0Q17Mt0eC76AAHElrsyusqGs BBaoP2RoyQDIdyTJaEe 8zQfRyJpi 4pEcgQ9XGfS7jy55g

It is not of the form .

Hence, the given equation is not a quadratic equation.U7LVZK8 JP0

It is of the form .

Hence, the given equation is not a quadratic equation.
qDvBEWWdXBrdZSrFafhTlFlP pTjdgOjMK ltc9OT0q6xIV2i7UXRd1JtrNcjb8fiTcT9s3tyvoBDED98xrjzTEptrDCnAxqUY7lpkWQVB6SE8kx RPDW0u9vEZtS 9DazKY

It is of the form

Hence, the given equation is a quadratic equation

3B0vgbwcRgaWAx6iFu6sM FyPNusGlJBrzyf7vc5RfQoYvojEdVSHk7w9b1MQnwibdQ78P UzHbSPU3iKMlkajTvQU6DZw3kgStdU5f5pyi4qcTo4l UhzcxJYv l0oZe yiFYM

  It is not of the form .

  Hence, the given equation is not a quadratic equation.

    It is not of the form

Hence, the given equation is not a quadratic equation.

hflg3xTDrhcpkj4A05oLWoWdqx eoGMomJe TRqUcHqi3QMVGxId6 rtJCjFyOeNV0xNd4yUDB mPQg5yosZDf56hgrJL hGZvCr0OvJWTz8rNBHHt

   It is of the form

Hence, the given equation is a quadratic equation.

Q2 :

Represent the following situations in the form of quadratic equations.

  1. The area of a rectangular plot is 528 m2. 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.
  2. The product of two consecutive positive integers is 306. We need to find the integers.
  3. 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.

  1. 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.

Answer :

  1. Let the breadth of the plot be x m. Hence, the length of the plot is (2x + 1) m. Area of a rectangle = Length x Breadth
10HRSUZzbGzlvjXVlK bVzYjP52sGUx94Sdug21YjidyZY9l RaEtfVXRt707q9HLl7Fp8a 6irEcvZHc0Btop3MMFBZrMrnqbnyJDPwRqRpW WgKnUbsmjSzmfaY7ajfLstcK8

∴ 528 = x (2x + 1)

  1. Let the consecutive integers be x and x + 1. It is given that their product is 306.
47 eDUT70XLR8YLFKdkd2w3v6w6Y1hv it7EnTHw0nRfxfjGR32xCutoyCJ3zjXwl3KbDJg nqpm6L6U8c0 vaTGgXWCh23njiasSREBLW0gwL93sDRNGeihLCnEpMDA3dDZU
  1. 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.
2ZoQOB1NDXq02PYo0pf1knw0PquU8tLxBoTSps4z53OwNvUXtgXjMQCL8WJ1Ay0B98rvaxxrnhn hyKQ2AD2nzUIvcT5s21AAAf48vpGbsSPF2DqN7bMKQNnpYLd3owbjj1FMA

  1. Let the speed of the train be x km/h. 

Time taken to travel 480 km =

In second condition, let the speed of train =   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 hrs Speed x Time = Distance

A7oCLl502zKZH
Y 8kGfzzMRXXpoz08cbATt ZnY xjnbWNqZc383yroh4iYGz7bOJDKj

Exercise 4.2 

Q1 

Find the roots of the following quadratic equations by factorisation:SXck35doF5AuXmAEf5gCaw HqVwEqbkmHThSgk5Z 8BP6a5N8d K5VU7aAnUfpS3ODw3VOHKcotqMS bkq7tuKS5i5k W6v8Dugiu2Imf6U2qhyYIWblJneA6wY6LA4Z d4D1kQ

Answer :

Roots of this equation are the values for which WRnOwqEqXpaHd8v pcOb5dX iWg68ehSeRGdQ vipOz7

∴  IlIdj4rdvlj10AEjjKrSIUqIww4umR59GGEltptXsz bqS0wiubG2SqWX75uZpLgJoxrL7srAcOG2FRAqgwUETfcvb0GALsi0Fo3CNAD wBfZNw7 ri3PNBViegeLWIgXLJ4c8= 0 or iPBRh snBfMp2 blTYQQ4T PjnSigFzvTZ0zQNA2XNiCnWpREBB0feF4sqzRjG TOLR 002MjUnlk4Av2vZLjp aDeATcUL=  i.e., x = 5 or x = – 2H exTM0Nn7E92aXV4MMGpjXFFB1EcSoMG50x0SKHTo0bEc9FrPr7DRpwVJTNu JMit0jp0yNSaBP6ZCWylmP FvZwDHnPskEcYsJtn9orb1gFPN9BNi9fi9vZuFEbL0 Lt55k

Roots of this equation are the values for which 

∴  iPBRh snBfMp2 blTYQQ4T PjnSigFzvTZ0zQNA2XNiCnWpREBB0feF4sqzRjG TOLR 002MjUnlk4Av2vZLjp aDeATcUL= 0 or FhQwlBbJgajAuIsPMZrlmy2htFliULpTp jrk5hF8MkK8Dng d1I5ljOF5Mka4nHoMDd eWRrfSiONgcurUaB T70cGt uQHyOncewSycQmprGLu7qSRnitOOFJszX8HQq7EaE0= 0 i.e., x = – 2 or x =  7qVuvTGytcPiN5KZfGEkG6PoYaFOKP1acH FfjO5EigDysK7YqcmOyVqzTidyXtNKhQVUHOfmzEd6BUtjgiSko1SkC0zh3txU0K9PqlYpu7vvlMe fA Xe6KJPZ4P 0ur nTmKo

Roots of this equation are the values for which

∴  1ZnJROeTNWBgbYom116VTvuvCdUwxb8c6jK6SzVtP7PK ZdNzcu2DI0 8YRBjYYZReXYLrvuENeJAH0EXP4WQUlSGDB9bB2XU9sIzTccFWjJ2bpgEjCi58092 5djomwLvtqWRQ= 0 or E5lDHdXcBaqktUya4G4zrXnhkRTT EjnFTayMRqK2918NXQiWcceWYIvBQbY61etkwwlaM8Sug21Db 2GaxbastZwNjkb= 0

i.e., x =KQjowVpq5XKUnzE2oaaO4Y WdeqXM or x =KZqD2ExJmG UP3cJfkNJvZWysIZZ6cphZGWgIUOEpujwn3mKICHDAxc5XCcHYu9tgYtIwWPk9E7R4mRyfk4N16GpLuacnf0glP5bXlX zhFw9VlfHyCZi5Wy0t1SAIpvwBdmNRs

Roots of this equation are the values for which

Therefore,mfL7jMjIky5idSiSjv31tu9WlibTVoBjglU8Yh0NK5CzUhoj 3 Yhzxbx9PqaDGMjgztUBVOwdKRnX3uc6hEMNE9aXuhHaXbUfp9eeXz

i.e.,A220jKOXeIGmWC5aofKCOiIt1IQcTI80GWgyXTJbtss9f26NJvXX LRwlMKWH7i1bNNo0V5Djkr EoABqkH12MmsJHVsB9l k QbySohYy0NlPIn5XSP86EyiTwIKcqzqWHqsPwgR3rX8ZcMi4ZlKuypr4PQ7F cv7t TYY00STmr8AHcJecJXNZq8XPnmcFzzXQDNAFuMEzm9Nd3aDDDqx27x k8N3xIwCfcEp1wUOc5B0z

  Roots of this equation are the values for 

Therefore, C5Twc7zVgsrtOZJghasXt31fCwgLYA3pc2cvHVR98S5GYntakghdsJQ7nDwT9Q97cRQ5hqKL8Y8gLuiT3mHyMNuxqhepf4SJiz2u90f5PYNqMCeZiNS8rOPIRqjxQUBOA pAmKY

i.e.,

TvvNGJncTQzgGQ03eIVt3F9RcpYq0jrHogyFq67O5rsGxXve0KIxEclhOBVooUOtGsnyg46nop6619s 8vhsA7y8DWOH DVmDUriU6Pg 2aZ6z3luz0VZ88NY mDSO50wSm92Ns
Q2 

Solve the given situations mathematically.

  1. 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. Find out how many marbles they had to start with.
  2. 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. Find out the number of toys produced on that day.

Answer :

(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.

cfrFNxuzFcSaYSRMZU2 xk JFAKxFYUX70P0INRq 051ZYTpzn30LcAhBd2j999fZ3xotEHbd8j3UlkBlAo8ySw0JWdUSfMD8C5K

Either = 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.

∴ Cost of production of each toy = Rs (55 – x)

It is given that, total production of the toys = Rs 750

z kCizfl3uP elczPSWVDGog49qBZ7nM8ecbqAhnEE 64PwJrZ0kxWZYUHRc IR

Either

Hence, the number of toys will be either 25 or 30.

Q3 

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

Answer:

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.ZzXJxeMYcjL jJN79flCW11kQvotaKCIsdKZnS63TBbAmM P DTFIMwg1fVE37WbyQcQQiX8nCAYD7C1 ikb7ym vHktMBRT37CG9D9oWGGDFY7PJobDS9aCCfHgknFCsIJMlPM

Either  = 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.

*Q4 

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

Answer :

Let the consecutive positive integers be x and x + 1.

Either Since the integers are positive, x can 

only be 13.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

*Q5 

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

Answer :

Let the base of the right triangle be x cm. Its altitude = (x – 7) cm

oPTcP1stHRGAf1CZSkBjhV3rDJfJQKYNkIb26zh2AAqlFA68755fzmOI TBZHIK6K0FMmfHZzSduTgBKyhn4MQMJg09BoqU7TQobu 9l9oJcrsYJ1Car4pd89JvGhEPiRCRCmEg

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.

*Q6 

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.

Answer:

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.

GTQgKqPTUrIYkcfsKZjWq44ADn31fwdYl

Either 2x + 15 = 0 or x – 6 = 0, i.e., x =    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 × 6 + 3 = Rs 15

Exercise 4.3

Q1 :

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

Answer :7QULsCc4BslOaYeEAVKFx0PxQkD UHz1CiFSU0jzoYgaKwybilDSfWiRQJMIuSyNX8V0OJwilRz6AP SnOyxpDBovoDcu5KgyAxAO 4MUFVjqDGtwtgeJOqrUdeiTSrzr44gU5U

(ii)

0Ru7xMaESAcuOhqGQ61o0tdQWqJgf9WH33t8IUzUJZrHXS7ilTCqtqSXlUeCleAJ9UDNAhfKVG5lBapy3 aBfyQbXEPjNJu j5DTMUB5dYPubb96GkBfJgEZtG2cEZfAv67knzA
Sd7Zz0YBeDmAn2TZ K57hb9MgO7iC0Q7xDqClHWWgJQvwoCJhQeQl S2mf2tYe92qjTo
9QEyZjtL0zWTLvoA9Y7EVu4nXwv0z3FfeOFg D btNinRXim5iZWiiHxabHQ cWsvR1zlwQDrXND455

Adding both sides

There are two solution:

5L3zqLLproWrHhRoxAbOEfL93WArEFXA6ygmGzIZIbCInWsG nywdT0aqwR3I20rjh2uDhmbOd7wOnbcOSpjpwXUHsWMsioDUKJUcetJszjKabw4pE8FG77cy8JnsamwV 7 8Q

(iii)

X37lvsDuORKabAKX0XKEAQg9HPTYJpybhHDup73peSc2HYBWq9ehkNGbO1S54NJi2qaYUF3Aq5ZSvVLe6xxtR0QDVxCxA9TkGq 7Nco5bLcPVz
Gdtw5tTSm7iCbu4mnFNNfFfvmObOIhX35 EQnPKsv9Nbra1OR4lhvNm pbICr DcRCjXJtgaavILfX2KgH5jsFch2g6bbRbRbn2we rnjCUlxFjaFfmtxn7MZv Dou3bQ64jYM
sWqsmGBzHkco1XINtQA6RudOue30jo3zeznI8cAW 0Oj0dDLNqYV20O S0rZkuBfu5ERNr5uUytPrd5ScNrZwTf35JnnFGcSi2MLD 9o81zvtvHXkVEvcj80aCfKGiJ3yfCOfBE

Adding on both sides

qSKYiWiNVhAI1CYU0yaf2M3G H6RyEcCsM80hGrex18azRiA2x9w5FuTQUTuE0izOvh8HAl9kQsO2oB 9tLce0K8ANnRC1f1IAvkajJz9Gt9u8hZ7tttbeMIIuwnbJIRlTdO6w
fO McC4RO6uFVO
08ycW6CcjomogTGLPekR4OvaN75i39cGcIQUm 6arSdkAb7Sy2n84qj5IVh1xEUSWYbN2L8UCQ4ly1WPptySpEVFk7SApIrRfRClxY oFPAAO1mx4 OE PT7X8bWbUkOl5FBN3o
Q2 

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

Answer :

(i)

xTtWLWoaX5zSEerF9WTqVRxfW5MLCZ2E0ZqlKkpPpKSAn9IsSenn0TOkVqKVqoXcM0hML4bmKqzxJ3V mhtYNjPiGnauHTS1toX4fMzGqaswoBXFNlhbe6MqmmSIZmZ4K7 zOUk

Roots

or

or

(ii)

MM8AmB4mMDj2BPMsc7F Y nLnxWRsgmMu7gBYU2YD9NS tYWBYI2i2iY50yKn

Roots

20LIZb doAdnabAF9GEEA5XmRDhc3TBf8 sB ahGnE8wtHYPxLVnMBedIU himV 0a DE M0aDwvCUZf

Hence there are two solutions

and

grujBNVhAVzGsVjmQoAowRWOOisVjAPKt0mLhibLPYuNNbYvHeGi7ZsxRNtjyPAa4FFlj2kR5RECjBNEmHCS02WO4R01s
aAezou8 rHt0CZ7nMMpfH0SfRRQmKkvBlWyKrDEU8M1wI X8vI Vf9t5TQN3zKvOhcYgfG0G PQDrctSRacZw14Rvpoemxl1pO5GQAoHbMK4iSq6Os6i4
Q3 

Find the roots of the following equations:0156Gk8PJEn ikRtNHFX1D9HobcowY 0IvKETXWihMCcw2rJT4HgcAvW7eJHOCbnga01XCtQkCvfktR4HiEERYMExyn9Raxt VnYy5eLquQRUxlk5Ze cbw94T4zXQc7wB6X3E0

Answer :

efQz9mjgycT9Qij49GmsMzPGTyNcBak79z97qUrW 0gDUz2jyptS1HHnacpMoFDMoRmIR51NNACrMedjbbO67OZNPNrWEfDo6TCfVK3LF bCvQzvjOK3Wr2lUaXowwQSR IPYtw
xKaLQyl P K6Gw5X F2EF0mtdXjEihfwD2QRvy7mD 83 CjXgaBSco C5OY9gDhVHSvZtdDrgIGCgUs44xy MqpyFc PFl4 rXJntgrBSzZZU2JELi RLMVsMrPmrWGTYYhUVnA

Q4:

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is . Find his present age.

Answer :

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 isUVHWjB8gs1wi 0x42 XePYgzVXfEhzA 2xhRugx4pdUmbKTvPYWC3QFYRzgi7gxs5tNCk9kfdEoKHQxl14OLMeS56IAyXpG .

dzni

However, age cannot be negative. Therefore, Rehman’s present age is 7 years.

Q5: 

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.

Answer :

Let the marks in Maths be x.

Then, the marks in English will be 30 – x. According to the given question,

8c6U69V8NwKJcD0NwooVyJeFN9hEMNHlVnuHHnGdNniGyop3q iHd1SqEISAR6k8lcRsWHzpKO2acBKoOVGviQ4zqojvYdj1DL rUVuMdQPPfJAooWYswSvOjA6o0gEL9Rn j0g

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

Q6: 

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.

Answer :

Let the shorter side of the rectangle be x m. 

Then, larger side of the rectangle = (x + 30) m

IE7ScigWVHyXyDQO0b7Y7S4U5ahTN0IIMrF7uG 2lqkz60Pl9pgRGp0JUScohSA1Met6p652sXtlKDIq1qglihH545R0gwgOLZtJio7JSe3fnMGdg3ONcn7JTsPzEpzE aaNcKg

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

*Q7:

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.

Answer :

Let the larger and smaller number be x and y respectively. According to the given question,

t gnioWWmBBtbEDF5lGgYMlTZucdgGH1ZxApKTf6aGoAzxqf4nDues8sHBAqkFwJraWOihzvucRnhIMjVJoNqwt6J3EX1ncHmOVAwP 4GfzseAGNsz4GNTD zj WrwQFPqCUZAA

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.E3dz8

Therefore, the numbers are 18 and 12 or 18 and – 12.

*Q8: 

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.

Answer:

Let the speed of the train be x km/hr.

Time taken to cover 360 km

According to the given question,

QHWfEGSXLne dM nksHnsyJDWdP5vbHkiyi9uJRrhBCHZ2m3mFCoAXSjowvOfD2mYD40wP4VhL6Ce NrBOD8V1F2JEFZm0X Yqf8 Nww4T3bi

However, speed cannot be negative. Therefore, the speed of train is 40 km/hr.

*Q9: 
Two water taps together can fill a tank in 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.

Answer :

Let the time taken by the smaller pipe to fill the tank be x hr. Time taken by the larger pipe

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 hours by both the pipes together. Therefore,

64ExoMeW 1SX2E79WSCH4TNg33K6wDfQcTKEz3SsYffRpHl psdrXvXTfn QAEG6OE13sCSbTPB83NYvZ8KG1E6ouPt mHNSF96k3TdUTD bSz3REsiCuZhBYFd3CWx 0Hwbeg

Time taken by the smaller pipe cannot be = 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.

*Q10: 

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 speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer :

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.

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.

*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.

Answer :

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 x2 and y2 respectively.

It is given that 4x – 4y = 24

x – y = 6

x = y + 6

nsFuYC3u9FbU1ExRL416EHvXzYkp2IRd Gvu Ido2yOQJ1v6HdfJk0s7LVowf48JStIRvWzKgSC6V7heSMoXKDpdTJr6X oibsqEb8jfYeOXvrPXiRT0hI6U40W1lMXBJ7k1FBE

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

Exercise – 4.4 

Q1

Find the nature of the roots of the following quadratic equations. If the real roots        exist, find them;

(I)

(II)

(III)

Answer :

We know that for a quadratic equation discriminant is b2 – 4ac.

If two distinct real roots

If two equal real roots

If no real roots 

(I)

Comparing this equation with ,  we obtain

mpvtFxEecxI8MNWdulKlR8Ueo8mbUCy7NG U68o8wiQ6BBNHglh9Aahqra3yl 0IRPbm7UUv508NhvRnE04CorG4kcSV437G9PE1L7OnExsgkAlWJ5gkh2RspOML5b D cvCr0A

Discriminant

= – 31

As b2 – 4ac < 0,

Therefore, no real root is possible for the given equation.

(II)

Comparing this equation with we obtain

MVQ1q2h1aHmXYIyMu0BeMQIT mVCl6Wdy3fLgygPxsxk CsbO rnNggaZ4QNR9 hHtQwaVPZbxzlINc8 hIk8jsz M8CF6 whWl7tx

Discriminant

= 48 – 48 = 0

As b2 – 4ac = 0,

Therefore, real roots exist for the given equation and they are equal to each other. And the roots will be

Therefore, the roots are

(III)

Comparing this equation with we obtain

a = 2, b = – 6, c = 3

Discriminant

= 36 – 24 = 12

As b2 – 4ac > 0,

Therefore, distinct real roots exist for this equation as follows.

yoXCKqkuczh40cd9vka0yEa8NKUhU w5Nc AVR45bUdhz4RlEEbMzF6q1qmGd EWKV MC18Of 7QiVXvZIUhfAquWkEl5hsMfr8EaBfbw4VSkMh6HQ8o39VQX YXuU L9QSy4s

Therefore, the roots are

Or

*Q2: 

Find the values of k for each of the following quadratic equations, so that they have two equal roots. 

(I)

(II) kx(x – 2) + 6 = 0

Answer :

We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant (b2 – 4ac) will be 0.

(I) 2x+kx + 3 = 0

Comparing equation with ax2 + bx + c = 0, we obtain

a = 2, b = k, c = 3

Discriminant

= k2 – 24

For equal roots, Discriminant = 0

Mh8spXD7CDWRyQ4al2TpS eGn8mCT5btlrymLZjtPwZqJq6A dgm wQIt43Qku62ZneKN 2xgZzyMjooq4KEkavZxShNDdiB6sbIjDayl8f3zw4kNH18QdwEI1BJNmf RiKJBgw

(II) or

0Yz8xN6sTes bE91rABV Nuwzw1gHEmcBaTRZkWH Vr27WGpnjltQzTG3Ju1emErWl1M

Comparing this equation with we obtain

UqX49klfxxjmXfb2TQKnyQZzTqXY

Discriminant

pnL95XtQYYa47Mt7isu wCVQTVLkJCmjzqCaVaI3gS0hLcwIJIKKeQzqF15PxVYBFdtOmVm1rHKJxu1rTAUNZbAypg PH9 Js2lLPAxWXbWFUkn1X1BszKwpV8AvxJlA0Srf mk

For equal roots,

8r Ose6g BgkUOCNzSGWvDPxhhYHl374a3GfFkWXzMj6a0jOp2nv0SlGRAYzQWcs2vBJaY byL9qW3hkSP5PrBQk6WOEVBZvMmSQe4XRuP3yUo9dgGMmZ2lRmraCHXs4n0Rrtc
zNCyF8I9l AtbCuU78AMXkKCD3XValRt0e XHemjxXwM UsEHcrEaLXfyJaIPdybgv AFSUAg7ld8ehTng7zmqGuHeTOgNtWQEe9q 2CApCUaNQxfZk1ktleh5p7hIO8NyL5 sk
qjBTdz87Kxjkfd 12IFgOMJ0yb3aYiaSJ 99buBnphZhZjjjOZAjHxyewA80KDUb82vSmenQ5MsVCuPuNH8IyOuF pphl1Ld20RJ9PMB02UuLdO1OsRdN

Either or

or

However, if k = 0, then the equation will not have the terms and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only.

*Q3 :

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Answer :

Let the breadth of mango grove be l.

 Length of mango grove will be 2l.

Area of mango grove = (2l) (l)

umKc kPGcUM JEwblrP HbH0e8q1u R1QVF6Mhq J5eG3 IYwBftfsgU4PSCrf PUGlWghJA gA

cpGsdgxUrOI t2aib1ZxMQftddZh6hf4ciPfrpqDC4ogHLta9UXdNceISNHkpxJvoLwM7bjVQmlsW8 KuoIylTQF8xJzV2nEHMeRTat02

Comparing this equation with we obtain

Discriminant Here,

Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

However, length cannot be negative. 

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

*Q4: 

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.

Answer :

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 

And, age of 2nd friend = (20 – x – 4)= (16 – x) years

Given that,

(x – 4) (16 – x) = 48

16x – 64 – x2 + 4x = 48

-x2 + 20x – 112 = 0

x2 – 20x + 112 = 0

Comparing this equation with ax2 + bx + c = 0, we obtain

a = 1, b = -20, c = 112

Discriminant

fQnRgCIxCrhadt6I CH5kvmqZtiulIyClikUsU46vzDnA0DAE wIhGFoB 2ycT fXPxtlwBXSTm1EMuoKJVTCZsNZlUAW8s qpoHCoWnEmRsvP5dHrIyj9C2ZZzkWh9GL645 RU

As

Therefore, no real root is possible for this equation and hence, this situation is not possible.

*Q5:

Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Answer :

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

KERPF uqwE6rZd0Vz2Sm9PVFeRkDpkobCnYkXPuOIFF pTfXPIDESBOE G3Mg1FUkvE0ySRYx94Ys l80r7 Gcl8tHUkE9Ea1J57o8SRY64dSxHR89IQ38ThznTmr7 qoetVR4
Z3A1Zh6stK1zPxzf jVvxJXWoo3cnzf

Comparing this equation with

we obtain

4PxRSxvMTxzHGPLjdSSE83aL4hArKLaZS0PCLLWcr34

Discriminate

zzm6U6ENYxLLjH605mCnEsbMR7oA8xeA xITp9m5n2K8ot3PaZJYjnLDedBM59OY9Mue uhNiJDICd 3d4G1mjcMhW32wYyk pS ujiG3YeCBULeMPWwD3O556Xj aPYuaj7oRM

As

Therefore, this equation has equal real roots.

And hence, this situation is possible. Root of this equation,

JCV5P1fervZzIZ8tMPL6oyUlBBh0fARATuRyiz4qcjNtFEfa6C yVj8U93 FkNK6eS9yqbeFtIBISY E634TJ0mWBOmQwYKCCrIqfPWP89gpFEl3vhv4DksgkyURYm fk2Y78SI

Therefore, length of park, l = 20 m

And breadth of park, b = 40 – l = 40 – 20 = 20 m

NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations
A 1 mark question was asked from Chapter 4 Quadratic Equations in the year 2018. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. Therefore, students need to have a thorough understanding of the topic. The topics and sub-topics provided in this chapter include:

4.1 Introduction If we equate the polynomial ax2+ bx + c, a ≠ 0 to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. In this chapter, students will study quadratic equations and various ways of finding their roots. They will also see some applications of quadratic equations in daily life situations. 4.2 Quadratic Equations A quadratic equation in the variable x is an equation of the form ax2+ bx + c = 0, where a, b, c are real numbers, a ≠ 0. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax2+ bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation. 4.3 Solution of Quadratic Equations by Factorization A real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2+ bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2+ bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. 4.4 Solution of a Quadratic Equation by Completing the Square Finding the value that makes a quadratic equation a square trinomial is called completing the square. The square trinomial can then be solved easily by factorizing. 4.5 Nature of Roots If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax2 + bx + c = 0 has (i) two distinct real roots, if b2 – 4ac > 0, (ii) two equal real roots, if b2– 4ac = 0, (iii) no real roots, if b2– 4ac < 0. 4.6 Summary
List of Exercises we covered in NCERT Solutions for Class 10 Maths Chapter 4:
Exercise 4.1 Solutions– 2 Questions
Exercise 4.2 Solutions– 6 Questions
Exercise 4.3 Solutions– 11 Questions
Exercise 4.4 Solutions– 5 Questions In a quadratic equation, x represents an unknown form and a, b, c are the known values. An equation to be quadratic “a” should not be equal to 0. The equation is of the form ax2 + bx + c = 0. The values of a, b, and c are always real numbers. A quadratic equation can be calculated by completing the square. A quadratic equation has:

Two different real roots.
No real roots.
Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 4
How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 4?
The fourth Chapter of NCERT Solutions for Class 10 Maths has 4 exercises. The first exercise deals with the topic of determining the quadratic equations, second exercise has questions of finding the roots of quadratic equations by factorization, third exercise deals with finding the roots of quadratic equations by completing squares and last exercise has the question based on the nature of the roots. By solving these exercises students are able to answer all the questions based on quadratic equations.
Is SWC’S website providing answers for NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations?
Yes, you can avail the PDFs of NCERT Solutions for Class 10 Maths Quadratic Equations. These solutions are formulated by expert faculty at SWC’S in an unique way. Also they provide solutions for Class 1 to 12 NCERT Textbooks in free PDFs. Who wish to score high in exams are advised to solve the NCERT Textbook.
Mention the important concepts that you learn in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations?
The concepts presented in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations are meaning and definition of quadratic equations, finding the roots of quadratic equations by factorization, finding the roots of quadratic equations by completing squares and nature of the roots.


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