**1.**

**Check whether the following are quadratic equations:**

**Solution:****(i) (x + 1)**^{2}= 2(x – 3)x^{2}+ 2x + 1 = 2x – 6x^{2}+ 7 = 0This is of the form ax^{2}+ bx + c = 0\ It is a quadratic equation.(ii) x^{2}– 2x = -2(3 – x)x^{2}– 2x = -6 + 2xx^{2}– 2x -2x + 6 = 0x^{2}– 4x + 6. This is of the form ax^{2}+ bx + c = 0It is a quadratic equation.(iii) (x - 2)(x + 1) = (x - 1)(x + 3)x^{2}– 2x + x - 2 = x^{2}+ 3x – x - 3– x – 2x +1 = 0- 3x + 1 = 0It is not of the form ax^{2}+ bx + cIt is not a quadratic equation.

**2.**

**Check whether the following are quadratic equations:**

**(i) (x – 3) (2x + 1) = x (x + 5)**

(ii) (2x – 1) (x – 3) = (x + 5) (x – 1) (iii) x

^{2}+ 3x + 1 = (x – 2)^{2}**Solution:****(i)(x – 3) (2x + 1) = x (x + 5)**2x^{2}+ x – 6x – 3 = x^{2}+ 5xx^{2}- 3 = 0This is of the form ax^{2}+ bx + c = 0Hence it is a quadratic equation.(ii)(2x – 1) (x – 3) = (x + 5) (x – 1)2x^{2}– 6x – x + 3 = x^{2}– x + 5x – 52x^{2}– 7x + 3 - x^{2}–4x + 5 = 0x^{2}– 11x + 8 = 0This is of the form ax^{2}+ bx + x = 0Hence it is a quadratic equation.(iii) x^{2}+ 3x + 1 = (x – 2)^{2}x^{2}+ 3x + 1 = x^{2}– 4x +47x – 3 = 0This is not of the form ax^{2}+ bx + c = 0\ It is not a quadratic equation.(iv) (x+2)^{3}= 2x(x^{2}– 1)x^{2}+ 2^{3}+ 3x^{2}´ 2 + 3x ´ 4 = 2x^{3}– 2xx^{3}+ 8 + 6x^{2 }+ 12x = 2x^{3}– 2xx^{3}– 6x^{2}– 14x – 8 = 0Its not of the form ax^{2}+ bx + c = 0It is not a quadratic equation.(v) x^{3}– 4x^{2}– x + 1 = (x – 2)^{3}x^{3}– 4x^{2}– x + 1 = x^{3}– 8 - 3x^{2}´ 2 + 3x ´ 2^{2}x^{3}– 4x^{2}– x + 1 = x^{3}– 8 – 6x^{2}+ 12x2x^{2}–13x + 1 = 0This of the form ax^{2}+ bx + c = 0Hence it is a quadratic equation.

**3.**

**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.**Solution:**(i)^{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**.**Let l be length (in metres) and b be the breadth (in metres) of the rectangle.Givenl = 1 + 2b …………………(1)lb = 528 m^{2}----- (2)

(Area of rectangle = length ´ breadth)Substituting (1) in (2)(1+2b)b = 528b +2b^{2}l = 5282b^{2}+ b – 528 = 0This is the quadratic equation where the breadth is in metres.(ii) The product of two consecutive positive integers is 306. We need need to find the integersLet the 2 consecutive positive integer be x, x + 1x(x + 1) = 306x^{2 }+ x - 306 = 0 is the quadratic equation where x is the smallestpositive integer

**4.**

**Represent the following situations in the form of quadratic equations:**

(i) 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.

**Solution:**(i) 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.Let the Rohans present age be xRohans mothers present age = x + 26After 3 years,Rohans age = x + 3Rohans mothers age = x + 26 + 3= x + 29(x + 3)(x + 29) = 360 (given)x^{2}+ 3x + 29x + 87 = 360x^{2}+ 32x – 273 = 0 where x is Rohans present age (in years)(ii) 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.Let the speed of the train be x km / hr\ Time taken =If speed = (x – 8) km/ hTime taken for the same distance = hrGiven 9; - = 3480x – 480 (x – 8) = 3(x – 8)x480x – 480x + 480 x 8 = 3x^{2}– 24x3x^{2}– 24x – 3840 = 0x^{2}– 8x - 1280 = 0 is the required quadratic equation where x is the speed of the train in km / hr.

**5.**

**Find the roots of the following quadratic equations by factorization:**

(i) x

(iii) + 7x + 5 = 0 (iv) 2x

^{2 }– 3x – 10 = 0 (ii) 2x^{2}+ x – 6 = 0(iii) + 7x + 5 = 0 (iv) 2x

^{2}– x + = 0**Solution:****(i)x**^{2 }– 3x – 10 = 0**x**^{2 }– 5x + 2x - 10 = 0x(x – 5) + 2(x – 5) = 0(x – 5)(x + 2) = 0\ x = -2, x = 5Thus –2, 5 are the roots of the equation.(ii) 2x^{2}+ x – 6 = 02x^{2}+ 4x – 3x – 6 = 02x(x + 2) – 3(x + 2) = 0(x + 2) (2x - 3) = 0Þ x = -2 (or) x = are the roots of the equation(iii) + 7x + 5 = 0x^{2 }+ 5x + 2x + 5 = 0x() + (x + 5) = 0(x + )(x + 5) = 0\ x = -, x = -\ The roots are -, -.(iv) 2x^{2}– x + = 016x^{2}– 8x + 1 = 016x^{2}– 4x – 4x + 1 = 04x(4x – 1) – (4x – 1) = 0(4x –1)^{2}= 0x = ,\ The root of the equation are , .(v) 100x^{2}– 20x + 1 = 0100x^{2}– 10x – 10x + 1 = 010x(10x – 1) –1 (10x – 1) = 0(10x – 1)^{2}= 0x = ,\ The roots of the equation are, .

**6.**

**Solve the following situation 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.

**Solution:**(i) Let the number of marbles John had be xThen the number of marbles Jivanti had = 45 – x.The number of marbles left with John, when he lost 5 marbles = x – 5The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5= 40 – xTherefore, their product = (x – 5)(40 – x)=40x – x^{2}– 200 + 5x= -x^{2}+ 45x – 200So, -x^{2}+ 45x – 200 = 124i.e., -x^{2}+ 45x – 324 = 0i.e., x^{2}- 45x + 324 = 0Therefore, the number of marbles John had, satisfies the quadratic equationx^{2}– 45x + 324 = 0Which is the required representation of the problem mathematically.Now by solving we get, x^{2}– 9x – 36 x + 324 = 0x(x – 9) - 36(x – 9) = 0(x - 9)(x - 36) = 0\ x = 9 (or) 36.\ The number of marbles John had is 9 (or) 36.(ii) Let the number of toys produced on that day be x.Therefore, the cost of production (in rupees) of each toy that day = 55 – xSo, the total cost of production (in rupees) that day = x(55 – x)Therefore, x(55 – x) = 750i.e., 55x – x^{2}= 750i.e., -x^{2}+ 55x – 750 = 0i.e., x^{2}- 55x + 750 = 0Therefore, the number of toys produced that day satisfies the quadratic equationx^{2}– 55x + 750 = 0which is the required representation of the problem mathematically.Now by solving we get, x^{2}– 25x – 30x + 750 = 0x(x – 25)- 30(x – 25) = 0(x - 25)(x - 30) = 0x = 25, x = 30.The number of toys produced in a day = 25 (or) 30.

**7.**

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

**Solution:**Let the two numbers be x and yx + y = 27 …………………….. (1)xy = 182 …………………….. (2)from (1) x = 27 – ySubstituting in (2) we get,(27 – y)y = 18227y – y^{2}= 182y^{2}– 27y + 182 = 0y^{2}– 13y - 14y + 182 = 0y(y – 13) – 14(y – 13) = 0(y – 14)(y - 13) = 0y = 14 (or) y = 13If y = 14, x = 27 – 14= 13If y = 13, x = 27 – 13= 14Hence the two numbers are 13 and 14.

**8.**

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

**Solution:**Let the 2 consecutive positive nos be x, x+1x^{2}+ (x + 1)^{2}= 365x^{2}+ x^{2}+ 2x + 1 = 3652x^{2}+ 2x – 364 = 0x^{2}+ x – 182 = 0 (Dividing by 2)Þ x^{2}+ 14x – 13x – 182 = 0x(x + 14) – 13(x + 14) = 0(x – 13)(x + 14) = 0Þ x = 13 (or) x = -14.Since the positive integers cannot be -14Hence the positive integers are 13 and 14.

**9.**

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

**Solution:**In the right angled DLet (BC) the base be x cm and altitude (AB) be (x – 7) cmIn the right angled D ,(AB)^{2}+ (BC)^{2}= (AC)^{2}(Pythagoras theorem)(x - 7)^{2}+ x^{2}= 13^{2}x^{2}– 14x + 49 + x^{2}= 1692x^{2}– 14x – 120 = 0x^{2}– 7x – 60 = 0 (Dividing by 2 )x^{2}– 12x +5x - 60 = 0x(x – 12) + 5(x – 12) = 0(x – 12)(x + 5) = 0=> x = -5 (or) x = 12Since the side (base) cannot be negative, x ¹ -5We assume x = 12=> Base (BC) = 12 cmAltitude = x – 7 = 5 cmThus the two sides of the triangle are 5 cm and 12 cm.

**10.**

**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 be x and the cost of each article be yGiven:y = 2x + 3 ………………..(1)Total cost of production be xy = 90 …………….. (2)Substuting y = 2x + 3 in (2), we getx(2x+3) = 902x^{2}+ 3x – 90 = 02x^{2}+ 15x – 12x – 90 = 02x^{2}+ 15x – 12x – 90 = 0x(2x + 15) - 6(2x + 15) = 0(2x + 15) (x – 6) = 02x + 15 = 0 ; x – 6 = 0=>x = ; 6Since the number of articles cannot be negative as well as a fraction x ¹Hence the number of articles is 6 (i.e.,) x = 6y = 2x + 3 from (1)= 2(6) + 3 = Rs. 15(i.e.,) Number of articles = 6cost of each article = Rs. 15**Solution:****(i)2x**^{2}– 7x + 3 = 0Dividing by the coefficient of x^{2}, we getx^{2}-Adding and subtracting the square of (coefficient of x) we get,x^{2}– 2 ´ ´ x+ = 0= 0= 0= 0Taking the square root we get,Whenx = =When(or) x =(ii) 2x^{2}+ x – 4 = 0Dividing by the coefficient of x^{2}, we getx^{2}+ = 0Adding and subtracting the square of (coefficient of x) we get,x^{2}+ 2(x) ´ += 0Taking the square root we get,x =\ x = ;(iii) 4x^{2}+4x + 3 = 0(2x)^{2}+ 2(2x) ´ + ()^{2}= 0(2x +)^{2}= 02x =-x = -\ Roots of equation are x = -(iv) 2x^{2}+ x + 4 = 0x^{2}+ x/2 + 2 = 0

Adding and subtracting the square of (coefficient of x) we get,x^{2}+ 2(x) ´= 0The square of a number cannot be negative .Hence roots do not exist.

**12.****Find the roots of the quadratic equations given in Q.1. above by applying the quadratic formula.**(i) 2x^{2}– 7x + 3 = 0 (ii) 2x^{2}+ x – 4 = 0**Solution:****(i) 2x**^{2}– 7x + 3 = 0According to the quadratic formula: ax^{2}+ bx + c = 0a = 2b = -7c = 3

\ The roots are === === = 3 ;The roots are 3, .(ii) 2x^{2}+ x – 4 = 0According to the quadratic formula: ax^{2}+ bx + c = 0a = 2, b = 1, c = -4\ The roots are==\ The roots are ,(iii) 4x^{2}+ 4 = 0According to the quadratic formula: ax^{2}+ bx + c = 0a = 4, b = 4, c = 3\ The roots are ===== ,The roots are =(iv)2x^{2}+ x + 4 = 0According to the quadratic formula: ax^{2}+ bx + c = 0a = 2, b = 1, c = 4\ The roots are ====This is not possible, hence the roots do not exists.

**13.****Find the roots of the following equations:****(i)****x ¹ 0****Solution:****(i)****x ¹ 0**x^{2}– 1 = 3xx^{2}– 3x – 1 = 0a = 1, b = -3, c = -1The roots are x =x ==(i.e.,) x = and(ii)(x+4)(x – 7) = -30x^{2}– 3x – 28 = -30x^{2}– 2x - x + 2 = 0x(x – 2) –1(x – 2) = 0(x – 2) (x – 1) = 0x = 1 (or) 2 are the roots of the equation.

**14.****The sum of the reciprocals of Rehman’s ages(in years) 3 years ago and 5 years from now is Find his present age.****Solution:**Let Rehman’s present age be x years.His age 3 years ago = x – 35 years from now = x + 5=Taking LCM,3(2x + 2) = (x – 3) (x + 5)6x + 6 = x^{2}+ 2x -15x^{2}– 4x – 21 = 0x^{2}– 7x + 3x - 21 = 0x(x – 7) + 3(x – 7) = 0Þ x = 7 or x = -3Age cannot be negative thus x = 7.\ His present age = 7 years

**15.****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:**Let her marks in Maths = x and in English = yGiven x + y = 30 …………………. (1)Given, (x + 2)(y – 3) = 210 …… (2)From (1) : x = 30 - y ……(3)Substututing in (2) we get(30 – y + 2) (y – 3) = 210(32 – y) (y – 3) = 21032y – 96 – y^{2}+ 3y = 210y^{2}– 35y + 306 = 0y^{2}– 17y – 18y + 306 = 0y(y – 17) – 18(y – 17) = 0(y – 18) (y – 17) = 0y = 18 (or) 17Þ x = 13 (when y= 17 by substituting in (3))x = 12 (when y= 18 by substituting in (3))Her marks in Maths is 13 and Science is 17 (or) Maths is 12 and Science is 18.

**16.****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:**Let x be the length of the shorter sideLonger side = (x + 30)In the right angled D BCD,(BD)^{ 2}= (BC)^{2}+ (CD)^{2}(Pythogoras theorem)(60 + x)^{2}= (x + 30)^{2}+ x^{2}(x + 60)^{2}= x^{2}+ x^{2}+ 60x + 9002x^{2}+ 60x + 900 = x^{2}+ 120 x + 3600x^{2}- 60x – 2700 = 0x^{2}- 90 x + 30x – 2700 = 0x(x - 90) + 30 (x - 90) = 0(x + 30)(x – 90) = 0x = -30; x = 90Since the length of a side cannot be negative , x = 90\ Shorter side = 90 m\ Longer side = 90 + 30 = 120 m

**17.****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 two numbers be x, yx^{2}– y^{2}= 180 ………………. (1)y^{2}= 8x ………………. (2)Substitute (2) in (1)x^{2}– 8x = 180x^{2}– 8x – 180 = 0x^{2}+ 10x – 18x – 180 = 0x(x + 10) – 18(x + 10) = 0(x + 10)(x – 18) = 0\ x = -10 ; x = 18If x = -10 from (2)y^{2}= 8x= 8(-10)= -80This is not possible as square of any cannot be negative.If x = 18 from (2)y^{2}= 8x= 8(18)= 144\ y = 12Thus the numbers are 18, 12 (or) 18, -12.

**18.****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 = x km/hrDistance traveled = 360 kmTime taken = =If the speed increases by 5km/hr the time required to reach the place reduces to one hourTime taken == 13603601800 = x^{2}+ 5xx^{2}+ 5x – 1800 = 0x^{2}+ 45x – 40x - 1800 = 0x(x + 45) – 40(x + 45) = 0(x + 45)(x – 40) = 0\ x = 40 (or) x = -45But speed of the train can only be positive, thus x = 40 km/hr

**19.****Two water taps together can fill a tank in 9 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:**Let the time taken by smaller tap = x hrs

In 1 hr of the tank can filled by smaller tap.Let the time taken by larger tap = (x – 10) hrsIn 1 hr of the tank can be filled by larger tap.&Together they fill the tank in==75(2x - 10) = 8(x^{2}– 10x)8x^{2}– 80x – 150x + 750 = 08x^{2}– 230x + 750 = 04x^{2}– 115x + 375 = 0 (Dividing by 2)4x^{2}– 100x – 15x + 375 = 04x(x – 25) – 15(x – 25) = 04x – 15 = 0 (or) x – 25 = 0x = (or) x = 25x = cannot satisfy the conditionSmall tap will fill the tank in 25 hrs and larger tap in 15 hrs.

**20.****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 the passenger train be x km/hr and the express train be (x + 11) km/hrTime taken by passenger train =Time taken by express train =According to the question,= 1**Solution:**Let the side of the two squares be a, b.Area of first square = a^{2}Area of second square = b^{2}\ a^{2}+ b^{2}= 468 …………….. (1)Perimeter of first square = 4aPerimeter of second square = 4b4a - 4b = 24Þ a – b = 6a = b + 6 ……………….. (2)substitute a in (1)(b+6)^{2}+ b^{2}= 468b^{2}+ 12b + 36 + b^{2}= 4682b^{2}+ 12b – 432 = 0Dividing by 2,b^{2}+ 6b – 216 = 0b^{2}+ 18b – 12b – 216 = 0b(b + 18) –12 (b +18) = 0(b + 18)(b – 12) = 0\ b = -18, b = 12Side cannot be negative thus b = 12 mSubstitute b = 12 in (2)a = b + 6= 12 + 6= 18 mSide of first square = 18 mSide of second square = 12 m

**22.****Find nature of roots of the following quadratic equation. If real roots exists find them.**

(i) 2x^{2}– 3x + 5 (ii) 3x^{2}- 4x + 4 (iii) 2x^{2}– 6x + 3 = 0**Solution:****(i) 2x**^{2}– 3x + 5Here a = 2, b = -3, c = 5Discriminant D = b^{2}– 4ac = 9 - (4 ´ 2 ´ 5)= 9 – 40= -31Equation 2x^{2}– 3x + 5 has imaginary roots.(ii) 3x^{2}- 4x + 4Herea = 3, b = -4, c = 4D = b^{2}– 4ac= (16 ´ 3) – (4 ´ 3 ´ 4)= (16 ´ 3) – (16 ´ 3)= 0The roots are real and equal.The roots are x =====,=,= ,(iii) 2x^{2}– 6x + 3 = 0a = 2 , b = -6, c = 3Discriminant D = b^{2}– 4ac= 36 – (4 ´ 2 ´ 3)= 36 – 24= 12D > 0\ Equation has real and distinct rootsx ===x =x = (or) x =

**23.****Find the value of k for each of the following quadratic equations, so that they have two equal roots.**(i) 2x^{2}+ kx + 3 = 0**Solution:****(i) 2x**^{2}+ kx + 3 = 0As the equation has equal roots the value of D = 0a = 2 , b = k, c = 3D = b^{2}– 4acZ = k^{2}– (4 ´ 2 ´ 3)0 = k^{2}– 24k^{2}= 24k =k = 2(ii) kx^{2}– 2kx + 6 = 0As the equation has equal roots the value of D = 0a = k , b = -2k, c = 6D = b^{2}– 4ac0 = 4k^{2}– (4 ´ k ´ 6)0 = 4k^{2}– 24k0 = k^{2}– 6kk(k – 6) = 0k = 0 (or) k = 6Thus k = 6.

**24.****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:**Let Length be x m and breadth be y mx = 2y (1)xy = 800 m^{2}substituting value of x from (1)2y ´ y = 8002y^{2}= 800=> y^{2}– 400 = 0y^{2}= 400y = 20 m Þ x = 40 mYes, it is possible to design a rectangular mango grove

**25.****Is the following situtation possible, If so determine the present ages.The sum of ages of 2 friends is 20 years. Four years ago, the product of their ages in year was 48.****Solution:**Let the present ages of friends be x, y.x + y = 20 (1)x = 20 –yThe product of their ages 4 yrs ago,(x – 4)(y – 4) = 48xy – 4y – 4x +16 = 48xy – 4y – 4x = 32(20 – y)y – 4y – 4(20 – y) = 3220y – y^{2}– 4y – 80 + 4y = 32-y^{2}+ 20y – 112 = 0D = b^{2}– 4ac= 400 – 4 ´ 1 ´ -112= 400 – 448= -48As D < 0 is not possible.

**26.****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:**Area = Length ´ Breadth = l ´ b = 400 m^{2}Perimeter = 2(l+b) = 80 m2(l + b) = 80l+b = 40 (1)b =

substituting value of b in equation (1) as .l + = 40l^{2}+ 400 = 40ll^{2}– 40l + 400 = 0D = b^{2}– 4ac= (-40)^{2}– (4 ´ 400 ´ 1)= 1600 – 1600= 0Equation has equal roots.As the equation has equal roots we can only design a square plot.

1321452 = x^{2}+ 11xx^{2}+11 x – 1452 = 0x^{2}+ 44x – 33x – 1452 = 0x(x + 44) – 33(x + 44) = 0(x – 33)(x + 44) = 0x = 33, x = -44Since the speed of train cannot be negative, x = 33 km / hrSpeed of the passenger train = 33 km/hrSpeed of the express train = 22km/hr