Exercise 4.2

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 .
        

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 (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.