Q.1    Check whether the following are quadratic equations :
          (i) (x+1)2=2(x−3)
          (ii) x2−2x=(−2)(3−x)
          (iii) (x−2)(x+1)=(x−1)(x+3)
          (iv) (x−3)(2x+1)=x(x+5)
          (v) (2x−1)(x−3)=(x+5)(x−1)
          (vi) x2+3x+1=(x−2)2
          (vii) (x+2)3=2x(x2−1)
          (viii) x3−4x2−x+1=(x−2)3
Sol.    (i) We have (x+1)2=2(x−3)
⇒ x2+2x+1=2x−6
⇒ x2+2x+1−2x+6=0
⇒ x2+7=0
Clearly, x2+7 is a quadratic polynomial. So the given equation is a quadratic equation.

(ii) We have , x2−2x=(−2)(3−x)
⇒ x2−2x+2(3−x)=0
⇒ x2−2x+6−2x=0
⇒ x2−4x+6=0
Clearly, x2−4x+6 is a quadratic polynomial. SO, the given equation is a quadratic equation.

(iii) We have , (x−2)(x+1)=(x−1)(x+3)
⇒ x2−x−2=x2+2x−3
⇒ x2−x−2−x2−2x+3=0
⇒ −3x+1=0
Clearly, −3x+1 is linear polynomial  So the given equation is not a quadratic equation.

(iv) We have, (x−3)(2x+1)=x(x+5)
⇒ x(2x+1)−3(2x+1)−x(x+5)=0
⇒ 2x2+x−6x−3−x2−5x=0
⇒ x2−10x−3=0
Clearly, x2−10x−3 is a quadratic polynomial. So, the given equation is a quadratic equation.

(v) We have, (2x−1)(x−3)=(x+5)(x−1)
⇒ (2x−1)(x−3)−(x+5)(x−1)=0
⇒ 2x(x−3)−1(x−3)−x(x−1)−5(x−1)=0
⇒ 2x2−6x−x+3−x2+x−5x+5=0
⇒ x2−11x+8=0
Clearly, x2−11x+8 is a quadratic polynomial. So, the given equation is a quadratic equation.

(vi) We have x2+3x+1=(x−2)2
⇒ x2+3x+1−(x−2)2=0
⇒ x2+3x+1−(x2−4x+4)=0
⇒ x2+3x+1−x2+4x−4=0
⇒ 7x−3=0
Clearly, 7x−3 is a linear polynomial. So, the given equation is not a quadratic equation.

(vii) We have, (x+2)3=2x(x2−1)
⇒ x3+3x2(2)+3x(2)2+(2)3=2x3−2x
⇒ x3+6x2+12x+8−2x3+2x=0
⇒ −x3+6x2+14x+8=0
Clearly, −x3+6x2+14x+8 being a polynomial of degree 3, is not a quadratic polynomial. So the given equation is not a quadratic equation.

(viii) We have, x3−4x2−x+1
=x3+3x2(−2)+3x(−2)2+(−2)3
⇒ x3−4x2−x+1=x3−6x2+12x−8
⇒ x3−4x2−x+1−x3+6x2−12x+8=0
⇒ 2x2−13x+9=0
Clearly, 2x2−13x+9 is a quadratic polynomial. So, the given equation is a quadratic equation.

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Q.2       Represent the following situation in the form of quadratic equations.
              (i) The area of a rectangular plot is 528m2. 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 year 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.

Sol.      (i) Let the length and breadth of the rectangular plot be 2x + 1 metres and x metres respectively. It is given that its area = 528m2.
Since (2x+1)×x=528
⇒ 2x2+x=528
⇒ 2x2+x−528=0,
Which is the required quadratic equation satisfying  the given conditions.

(ii) Let two consecutive integers be x and x + 1 such that their product = 306.
⇒ x (x + 1) = 306
⇒ x2+x−306=0
Which is the required quadratic equation satisfying  the given conditions.

 

(iii) Let Rohan ‘s present age be x years. Then,
His mother’s age = (x + 26) years.
After 3 years, their respective ages are (x + 3) years and (x + 29) years. … (1)
It is given that the product of, ages mentioned at (1) is 360
i.e. (x + 3)(x + 29) = 360
⇒ x2+32x+87=360
⇒ x2+32x+87−360=0
⇒ x2+32x−273=0
Therefore, the age of Rohan satisfies the quadratic equation x2+32x−273=0.

(iv) Let u km/hr be the speed of the train.
Then, time taken to cover 480 km =480uhours
Time taken to cover 480 km when the speed  is decreased. by 8 km/hr
=480u−8hours
It is given that the time to cover 480 km is increased by 3 hours.
Therefore, 480u−8−480u=3
⇒ 480u−480(u−8)=3u(u−8)
⇒ 160u−160u+1280=u2−8u
⇒ u2−8u−1280=0
Therefore, the speed of the train satisfies the quadratic equation u2−8u−1280=0

 

 

Q.1     Find the roots of the following quadratic equations by factorization :
           (i) x2−3x−10=0
           (ii) 2x2+x−6=0
           (iii) 2–√x2+7x+52–√=0
           (iv) 2x2−x+18=0
           (v) 100x2−20x+1=0
Sol.     (i) We have, x2−3x−10=0
⇒ x2−5x+2x−10=0
⇒ x(x−5)+2(x−5)=0
⇒ (x−5)(x+2)=0
⇒ x−5=0orx+2=0
⇒ x=5orx=−2
Thus , x = 5 and x = – 2 are two roots of the equation x2−3x−10=0.

(ii) We have, 2x2+x−6=0
⇒ 2x2+4x−3x−6=0
⇒ 2x(x+2)−3(x+2)=0
⇒ (x+2)(2x−3)=0
⇒ x+2=0or2x−3=0
⇒ x=−2orx=32
Thus x=−2orx=32 are two roots of the equation 2x2+x−6=0

 

(iii) We have, 2–√x2+7x+52–√=0
⇒ 2–√x2+2x+5x+52–√=0
⇒ 2–√x(x+2–√)+5(x+2–√)=0
⇒ (x+2–√)(2–√x+5)=0
⇒ x+2–√=0
or 2–√x+5=0
⇒ x=−2–√
or x=−52√=−52√2
Thus, x=−2–√andx=−52√2 are two roots of the equation 2–√x2+7x+52–√=0.

(iv) We have, 2x2−x+18=0
⇒ 16x2−8x+1=0
⇒ 16x2−4x−4x+1=0
⇒ 4x(4x−1)−1(4x−1)=0
⇒ (4x−1)(4x−1)=0
⇒ 4x−1=0
or 4x−1=0
⇒ x=14orx=14
Thus x=14orx=14 are two roots of the equation
2x2−x+18=0.

(v) We have , 100x2−20x+1=0
⇒ 100x2−10x−10x+1=0
⇒ 10x(10x−1)−1(10x−1)
⇒ (10x−1)(10x−1)=0
⇒ 10x−1=0
or 10x−1=0
⇒ x=110orx=110
Thus , x=110orx=110 are two roots of the equation 100x2−20x+1=0

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Q.2      Solve the problems given in Example 1 i.e., to solve
            (i) x2−45x+324=0
            (ii) x2−55x+750=0
            using factorisation method.
Sol.     (i) We have , x2−45x+324=0
⇒ x2−9x−36x+324=0
⇒ x(x−9)−36(x−9)=0
⇒ (x−9)(x−36)=0
⇒ x−9=0orx−36=0
⇒ x=9orx=36
Thus x = 9 and x = 36 are two roots of the equation x2−45x+324=0.

(ii) We have, x2−55x+750=0
⇒ x2−30x−25x+750=0
⇒ x(x−30)−25(x−30)=0
⇒ (x−30)(x−25)=0
⇒ x−30=0orx−25=0
⇒ x=30orx=25
Thus, x = 30 and x = 25 are two roots of the equation x2−55x+750=0.

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Q.3     Find two numbers whose sum is 27 and product is 182.

Sol.     Let the required numbers be x and 27 – x. Then x(27−x)=182
⇒ 27x−x2=182
⇒ x2−27x+182=0
⇒ x2−13x−14x+182=0
⇒ x(x−13)−14(x−13)=0
⇒ (x−13)(x−14)=0
⇒ x−13=0orx−14=0
⇒ x=13orx=14
Hence, the two numbers are 13 and 14.

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Q.4      Find two consecutive positive integers, sum of whose squares is 365.
Sol.      Let the two consecutive positive integers be x and x + 1.
Then , x2+(x+1)2=365
⇒ x2+x2+2x+1=365
⇒ 2x2+2x−364=0
⇒ x2+x−182=0
⇒ x2+14x−13x−182=0
⇒ x(x+14)−13(x+14)=0
⇒ (x+14)(x−13)=0
⇒ x+14=0orx−13=0
⇒ x=−14orx=13
Since x, being a positive integer, cannot be negative.
Therefore, x = 13.
Hence, the two consecutive positive integers are 13 and 14.

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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.
Sol.      Let the base of he right angled  triangle be x cm.
Then, its height is (x – 7) cm.
It given that the hypotenuse = 13
⇒ x2+(x−7)2−−−−−−−−−−−√=13
⇒ x2+(x2−14x+49)=169
⇒ 2x2−14x−120=0
⇒ x2−7x−60=0
⇒ x2−12x+5x−60=0
⇒ x(x−12)+5(x−12)=0
⇒ (x−12)(x+5)=0
⇒ x−12=0orx+5=0
⇒ x=12orx=−5
⇒ x=12
[Since, side of a triangle can never be negative]
Therefore, Length of the base = 12 cm and , Length of the height = (12 – 7) cm = 5 cm.

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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.
Sol.     Let the number of articles produced in a day = x
Then, cost of  production of each article = Rs (2 x + 3)
It is given that the total cost of production = Rs 90
Therefore, x × (2 x + 3) = 90
⇒ 2x2+3x−90=0
⇒ 2x2−12x+15x−90=0
⇒ 2x(x−6)+15(x−6)=0
⇒ (x−6)(2x+15)=0
⇒ x−6=0or2x+15=0
⇒ x=6orx=−152
⇒ x=6
[Since, the number of articles produced cannot be negative]
Cost of each article = Rs (2 × 6 + 3) = Rs 15
Hence, the number of articles produced are 6 and the cost of each article is Rs 15.

 

Q.1      Find the roots of the following quadratic equations, if they exist by the method of completing the square:
            (i) 2x2−7x+3=0
            (ii) 2x2+x−4=0
            (iii) 4x2+43–√x+3=0
            (iv) 2x2+x+4=0
Sol.      (i) The equation 2x2−7x+3=0 is the same as x2−72x+32=0
Now, x2−72x+32
=(x−74)2−(74)2+32
=(x−74)2−4916+32
=(x−74)2−2516
Therefore, 2x2−7x+3=0
⇒ (x−74)2−2516=0
⇒ (x−74)2=2516
⇒ x−74=±54
⇒ x=74±54
⇒ x=74+54=124=3
⇒ x=74−54=24=12
Therefore, The roots of the given equation are 3 and 12.

(ii) We have, 2x2+x−4=0
⇒ x2+x2−2=0
⇒ (x+14)2−(14)2−2=0
⇒ (x+14)2−116−2=0
⇒ (x+14)2−3316=0
⇒ (x+14)2=3316
⇒ x+14=±33√4
⇒ x=−14±33√4
⇒ x=−1+33√4
⇒ x=−1−33√4
Therefore, The roots of the given equation are −1−33√4 and −1+33√4
(iii) We have, 4x2+43–√x+3=0
⇒ (2x)2+2×(2x)×3–√+(3–√)2−(3–√)2+3=0
⇒ (2x+3–√)2−3+3=0
⇒ (2x+3–√)2=0
⇒ x=−3√2
⇒ x=−3√2
Therefore, The roots of the given equation are −3√2 and −3√2.

(iv) We have, 2x2+x+4=0
⇒ x2+12x+2=0
⇒ (x+14)2−116+2=0
⇒ (x+14)2+3116=0
⇒ (x+14)2=−3116<0
But (x+14)2 cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.

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Q.2      Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Sol.      (i) The given equation is 2x2−7x+3=0
Here a = 2, b = – 7 and c = 3.
Therefore, D=b2−4ac=(−7)2−4×2×3 = 49−24=25>0
So, the given equation has real roots given by
x=−b±D√2a
=−(−7)±25√2×2
=7±54
=124or24=3or12

(ii) The given equation is 2x2+x−4=0
Here, a = 2, b = 1 and c = – 4
Therefore, D=b2−4ac=(1)2−4×2×−4=1+32=33>0
So, the given equation has real roots given by
x=−b±D√2a=−1±33√2×2 =−1±33√4

(iii) The given equation is 4x2+43–√x+3=0
Here a = 4, b=43–√andc=3
Therefore, D=b2−4ac=(43–√)2−4×4×3=48−48=0
So, the given equation has real equal roots given by
x=−b±D√2a=−b2a =−43√2×4=−3√2
(iv) The given equation is 2x2+x+4=0
Here, a = 2, b = 1 and c = 4
Therefore, D=b2−4ac=(1)2−4×2×4=1−32=−31<0
So, the given equation has no real roots.
I prefer to use quadratic formula method as it is a straight forward method.

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Q.3      Find the roots of the following equations :
            (i) x−1x=3,x≠0
            (ii) 1x+4−1x−7=1130,x≠−4,7
Sol.       (i) The given equation is x−1x=3,x≠0
⇒ x2−3x−1=0
Here , a = 1, b = – 3 and c = – 1
Therefore, D=b2−4ac=(−3)2−4(1)(−1)=9+4=13>0
So, the given equation has real roots given by
x=−b±D√2a=−(−3)±13√2×1 =3±13√2

(ii) The given equation is 1x+4−1x−7=1130,x≠−4,7
⇒ (x−7)−(x+4)(x+4)(x−7)=1130
             ⇒x−7−x−4(x+4)(x−7)=1130
⇒ −11(x+4)(x−7)=1130
             ⇒−1x2−7x+4x−28=130
             ⇒ −1x2−3x−28=130
             ⇒ −30=x2−3x−28
⇒ x2−3x+2=0
            ⇒x2−2x−x+2=0
             ⇒ (x−1)(x−2)=0
             ⇒x(x−2)−1(x−2)=0
⇒ x=1or2
Thus x = 1 and x = 2 are the roots of the given equation.

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Q.4      The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 13. Find his present age.
Sol.      Let Rehman’s present age be x – years.
As per question, we have
1x−3+1x+5=13
⇒ 3(x+5)+3(x−3)=(x−3)(x+5)
⇒ 3x+15+3x−9=x2+2x−15
⇒ x2−4x−21=0
⇒ x2−7x+3x−21=0
            ⇒ (x−7)(x+3)=0
            ⇒x(x−7)+3(x−7)=0
            ⇒(x+3)(x−7)=0
⇒ x=7or−3
Therefore, x=7 [Since, age can never be negative]
Thus, Rehman’s present age is 7 years.

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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.
Sol.      Let Shefali’s marks in Mathematics be x. Then her marks in English will be (30 – x).
As per condition of the problem :
            (x+2)×[(30−x)−3]=210
⇒ (x+2)(27−x)=210
⇒ 27x−x2+54−2x=210
            ⇒ x2−25x+156=0
            ⇒ x2−12x−13x+156=0
            ⇒ x(x−12)−13(x−12)=0
            ⇒ (x−12)(x−13)=0
            ⇒ x=12orx=13
        Therefore, Shafali’s marks in Mathematics and English are 12 and 18 respectively or in Mathematics and English are 13 and 17, respectively.

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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.
Sol.      Let in the rectangular field BC = x metres. Then AC = (x + 60) and AB = (x + 30) metres. By Pythagoras Theorem, we have

           AC2=BC2+AB2
           ⇒ (x+60)2=x2+(x+30)2
           ⇒ x2+120x+3600=x2+x2+60x+900
           ⇒ x2−60x−2700=0
           ⇒ x2−90x+30x−2700=0
           ⇒ x(x−90)+30(x−900)=0
⇒ (x+30)(x−90)=0
⇒ x=−30orx=90
           ⇒ x=90 [Since side of a rectangle can never be negative]
           Hence, the sides of the field are 120 m and 90 m.

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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.
Sol.    Let the larger number be x. Then ,
          Square of the smaller number = 8x
          Also , square of the larger number = x2
          It is given that the difference of the squares of the number is 180
          Therefore, x2−8x=180
          ⇒ x2−8x−180=0
          ⇒ x2−18x+10x−180=0
          ⇒ x(x−18)+10(x−18)=0
          ⇒ (x−18)(x+10)=0
          ⇒ x−18=0orx+10=0
          ⇒ x=18orx=−10
          Case 1 : When x = 18. In this case, we have
          Square of the smaller number = 8x = 8 × 18 = 144
          Therefore, Smaller number =±12
          Thus, the number are 18, 12 or 18, – 12.
          Case 2 : When x = – 10
          In this case, we have
          Square of the smaller number = 8x = 8 × – 10 = – 80
          But , square of a number is always positive.
          Therefore, x = – 10 is not possible.
          Hence, the numbers are 18, 12 or 18 , – 12.

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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.
Sol.       Let x km/hr be the uniform speed of the train.
            Then time taken to cover 360km=360xhours.
            Time taken to cover 360 km when the speed is increased by 5km/hr=360x+5hours.
It is given that the time to cover 360 km is reduced by 1 hour.
Therefore, 360x−360x+5=1
            ⇒ 360(x+5)−360x=x(x+5)
            ⇒ 360x+1800−360x=x2+5x
            ⇒ x2+5x−1800=0
            ⇒ x2+45x−40x−1800=0
            ⇒ x(x+45)−40(x+45)=0
            ⇒ (x−40)(x+45)=0
⇒ x−40=0orx+45=0
⇒ x=40orx=−45
            But x cannot be negative. Therefore, x = 40.
            Hence, the original speed of the train is 40 km/hr.

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Q.9     Two water taps together can fill a tank in 938 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.
Sol.        Let the smaller tap takes x hours to fill the tank. So, the larger tap will take (x – 10) hours to fill the tank.
Therefore, Portion of the tank filled by the larger tap in one hours =1x−10
⇒ Portion of the tank filled by the larger tap 938hours
i.e., 758hours=1x−10×758
Similarly, portion of the tank filld by the smaller tap in 758hours=1x×758
Since it is given that the tank is filled in 758hours
Therefore, 758(x−10)+758x=1
⇒ 1x−10+1x=875
⇒ x+x−10x(x−10)=875
⇒ 75(2x−10)=8x(x−10)
⇒ 150x−750=8x2−80x
⇒ 8x2−230x+750=0
⇒ 4x2−115x+375=0
             ⇒ 4x2−100x−15x+375=0
             ⇒ 4x(x−25)−15(x−25)=0
             ⇒ (x−25)(4x−15)=0
             ⇒ x−25=0or4x−15=0
             ⇒ x=25or154
             Therefore, x=25asx=154 is inadmissible
             Hence, the larger tap fills the tank in 15 hours and the smaller tap takes 25 hours to fill the tank.

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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 11 km/h more than that of the passenger train, ,find the average speed of the two trains.
Sol.      Let the speed of the express train be x km/hr.
Then, the speed of the passenger train is (x – 11) km/hr
Distance to be covered by train = 132
Time taken by express train =132xhours and, time taken by the passenger train =132x−11
By the given condition, 132x−11−132x=1
⇒ 132x−132x+1452=x2−11x
⇒ x2−11x−1452=0
⇒ x2−44x+33x−1452=0
⇒ x(x−44)+33(x−44)=0
⇒ (x−44)(x+33)=0
⇒ x=44
or x=−33
But x cannot be negative. Therefore, x = 44
Therefore, Speed of express train = 44 km/hr
and, speed of passenger train = (44 – 11) km/hr = 33 km/hr.

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Q.11     Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Sol.       Let the sides of the squares be x and y metres (x > y).
According to question :
             x2+y2=468 … (1)
             and 4x−4y=24
             ⇒ x−y=6 … (2)
             Putting x = y + 6 in (1), we get (y+6)2+y2=468
            ⇒y2+12y+36+y2=468
⇒ 2y2+12y+36−468=0
            ⇒2y2+12y−432=0
⇒ y2+6y−216=0
            ⇒ y2+18y−12y−216=0
            ⇒ y(y+18)−12(y+18)=0
            ⇒ (y+18)(y−12)=0
            ⇒ y=−18ory=12
            But y cannot be negative. Therefore, y = 12
            Therefore, x = y + 6 = 12 + 6 = 18
            Therefore, The sides of the squares are 18 m and 12 m.

 

 

Q.1      Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
            (i) 2x2−3x+5=0
            (ii) 3x2−43–√x+4=0
            (iii) 2x2−6x+3=0
Sol.       (i) The given equation is 2x2−3x+5=0
             Here , a = 2, b = – 3 and c = 5
             Therefore, D=b2−4ac=(−3)2−4×2×5=9−40=−31<0
             So, the given equation has no real roots.

(ii) The given equation is 3x2−43–√x+4=0
Here a = 3, b=−43–√andc=4
Therefore, D=b2−4ac=(−43–√)2−4×3×4=48−48=0
So, the given equation has real equal roots, given by
x=−b±D√2a=−(−43√)±02×2=3–√.

(iii) The given equation is 2x2−6x+3=0
Here, a = 2, b = – 6 and c = 3
Therefore, D=b2−4ac=(−6)2−4×2×3=36−24=12>0
So, the given equation has real roots , given by
x=−b±D√2a=−(−6)±12√2×2
=6±23√4=3±3√2

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Q.2      Find the values of k for each of the following quadratic equations, so that they have two equal roots.
            (i) 2x2+kx+3=0
            (ii) kx(x−2)+6=0
Sol.      (i) The given equation is 2x2+kx+3=0
Here, a = 2, b = k and c = 3
Therefore, D=b2−4ac=k2−4×2×3=k2−24
The given equation will have real and equal roots, if
D = 0 ⇒ k2−24=0 ⇒ k=±24−−√=±26–√
(ii) The given equation is kx (x – 2) + 6 = 0
⇒ kx2−2kx+6=0
Here a = k , b = – 2k and c = 6
Therefore, D=b2−4ac=(−2k)2−4×k×6=4k2−24k
The given equation will have real and equal roots, if D = 0
⇒ 4k2−24k=0 ⇒ 4k(k−6)=0
⇒ k=0ork=6

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Q.3      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.
Sol.      Let 2x be the length and x be the breadth of a rectangular mango grove.
Area = (2x) (x) = 800 [Given]
⇒ x2=400
⇒ x=20 [Since cannot  be negative]
The value of x is real so design of grove is possible
Its length = 40 m and breadth = 20 m.

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Q.4      Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friend is 20 years. Four year ago, the porudct of their ages in years was 48.
Sol.      Let age of one of the friends = x years
Then, age of the other friend = 20 – x Four years ago,
            Age of one of the friend = (x – 4) years
            and age of the other friend = (20 – x – 4) years
            = (16 – x) years
            According to condition :
            (x−4)(16−x)=48
            ⇒ 16−x2−64+x=48
            ⇒ x2−20+112=0
            Here a = 1, b = – 20 and c = 112
            Therefore, D=b2−4ac=(−20)2−4×1×112
            =400−448=−48<0
            So, the given equation has no real roots.
            Thus , the given situation is not possible.

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Q.5      Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Sol.      Let length be x metres and breadth be y metres.
Therefore, Perimeter = 80 m
⇒ 2(x+y)=80
⇒ x+y=40 …(1)
Also, Area=400m2
⇒ xy=400
⇒ x(40−x)=400 [Using (1)]
⇒ 40x−x2=400
⇒ x2−40x+400=0
Here a = 1, b = – 40 and c = 400
Therefore,D=b2−4ac
=(−40)2−4×1×400
           =1600−1600=0
           So, the given equation has equal real roots.
           Therefore, Its length and breadth is given by
           x2−40x+400=0
           ⇒ (x−20)2=0
           ⇒ x=20,20
           Therefore, Length = 20 m
           Breadth = 20 m
           Therefore, Design is possible.