Exercise 4.3

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      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.6      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.7      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.8     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.9     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.