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Q1. Check whether the following are quadratic equations:
(i) (x + 1)^{2} = 2(x − 3)
(ii) x^{2} − 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) x^{2} + 3x + 1 = (x − 2)^{2}
(vii) (x + 2)^{3} = 2x(x^{2} − 1)
(viii) x^{3} − 4x^{2} − x + 1 = (x − 2)^{3}
Ans: (i) (x + 1)^{2} = 2(x−3)
We have:
(x + 1)^{2} = 2 (x − 3)
⇒ x^{2 }+ 2x + 1 = 2x − 6
⇒ x^{2} + 2x + 1 − 2x + 6 = 0
⇒ x^{2} + 7 = 0
Since x^{2} + 7 is a quadratic polynomial
∴ (x + 1)^{2} = 2(x − 3) is a quadratic equation.
(ii) x^{2}−2x = (− 2) (3 −x)
We have:
x^{2} − 2x = (− 2) (3 − x)
⇒ x^{2} − 2x = − 6 + 2x
⇒ x^{2} − 2x − 2x + 6 = 0
⇒ x^{2} − 4x + 6 = 0
Since x^{2 }− 4x + 6 is a quadratic polynomial
∴ x^{2} − 2x = (−2) (3 − x) is a quadratic equation.
(iii) (x−2) (x + 1) = (x−1) (x + 3)
We have:
(x − 2) (x + 1) = (x − 1) (x + 3)
⇒ x^{2} − x − 2 = x^{2} + 2x − 3
⇒ x^{2} − x − 2 − x^{2} − 2x + 3 = 0
⇒ −3x + 1 = 0
Since −3x + 1 is a linear polynomial
∴ (x − 2) (x + 1) = (x − 1) (x + 3) is not quadratic equation.
(iv) (x−3) (2x + 1) = x(x + 5)
We have:
(x − 3) (2x + 1) = x (x + 5)
⇒ 2x^{2} + x − 6x −3= x^{2 }+ 5x
⇒ 2x^{2} − 5x − 3 − x^{2} − 5x =0
⇒ x^{2} + 10x − 3= 0
Since x^{2} + 10x − 3 is a quadratic polynomial
∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.
(v) (2x−1) (x−3) = (x + 5) (x−1)
We have:
(2x − 1) (x − 3) = (x + 5) (x − 1)
⇒ 2x^{2} − 6x − x + 3 = x^{2} − x + 5x − 5
⇒ 2x^{2} − x^{2} − 6x − x + x − 5x + 3 + 5 = 0
⇒ x^{2} − 11x + 8 = 0
Since x^{2} − 11x + 8 is a quadratic polynomial
∴ (2x − 1) (x − 3) = (x + 5) (x − 1) is a quadratic equation.
(vi) x^{2} + 3x + 1 = (x−2)^{2}
We have:
x^{2} + 3x + 1 = (x − 2)^{2}
⇒ x^{2} + 3x + 1 = x^{2} − 4x + 4
⇒ x^{2} + 3x + 1 − x^{2 }+ 4x −4= 0
⇒ 7x − 3 = 0
Since 7x − 3 is a linear polynomial
∴ x^{2} + 3x + 1 = (x − 2)^{2 }is not a quadratic equation.
(vii) (x + 2)^{3} = 2x(x^{2}−1)
We have:
(x + 2)^{3} = 2x(x^{2} − 1)
⇒ x^{3} + 3x^{2}(2) + 3x(2)^{2} + (2)^{3} = 2x^{3} − 2x
⇒ x^{3} + 6x^{2} + 12x + 8 = 2x^{3} − 2x
⇒ x^{3} + 6x^{2} + 12x + 8 − 2x^{3} + 2x = 0
⇒− x^{3} + 6x^{2} + 14x + 8 = 0
Since −x^{3} + 6x^{2} + 14x + 8 is a polynomial of degree 3
∴ (x + 2^{)3} = 2x(x^{2} − 1) is not a quadratic equation.
(viii) x^{3}−4x^{2}−x + 1 = (x− 2)^{3}
We have:
x^{3} − 4x^{2} − x + 1 = (x − 2)^{3}
⇒ x^{3} − 4x^{2} − x + 1 = x^{3} + 3x^{2}(− 2) + 3x(− 2)^{2} + (− 2)^{3}
⇒ x^{3} − 4x^{2} − x + 1 = x^{3} − 6x^{2} + 12x − 8
⇒ x^{3} − 4x^{2} − x − 1 − x^{3} + 6x^{2} − 12x + 8 = 0
⇒ 2x^{2} − 13x + 9 = 0
Since 2x^{2} − 13x + 9 is a quadratic polynomial
∴ x^{3} − 4x^{2} − x + 1 = (x − 2)^{3} is a quadratic equation.
Q2. 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.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) 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.
(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.
Ans: (i) Let the breadth = x metres
∵ Length = 2 (Breadth) + 1
∴ Length = (2x + 1) metres
Since Length × Breadth = Area
∴ (2x + 1) × x = 528
⇒ 2x^{2} + x = 528
⇒ 2x^{2} + x − 528 = 0
Thus, the required quadratic equation is 2x^{2} + x− 528 = 0
(ii) Let the two consecutive numbers be x and (x + 1).
∵Product of the numbers = 306
∴ x (x + 1) = 306
⇒ x^{2} + x = 306
⇒ x^{2} + x − 306 = 0
Thus, the required equdratic equation is x^{2} + x − 306 = 0
(iii) Let the present age = x
∴ Mother’s age = (x + 26) years
After 3 years
His age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
=(x + 29) years
According to the condition,
⇒ (x + 3) × (x + 29) = 360
⇒ x^{2} + 29x + 3x + 87 = 360
⇒ x^{2} + 29x + 3x + 87 − 360 = 0
⇒ x^{2} + 32x − 273 = 0
Thus, the required quadratic equation is
x^{2} + 32x − 273 = 0
(iv) In first case,
Let the speed of the train = u km/hr
Distance covered = 480 km
Time taken = Distance ÷ Speed
In second case,
Speed = (u − 8) km/hour
According to the condition,
⇒ 480u − 480(u − 8) = 3u(u − 8)
⇒ 480u − 480u + 3840 = 3u^{2} − 24u
⇒ 3840 − 3u^{2} + 24u =0
⇒ 1280 − u^{2 }+ 8u =0
⇒ − 1280 + u^{2} − 8u =0
⇒ u^{2} − 8u − 1280 = 0
Thus, the required quadratic equation is u^{2} − 8u – 1280 = 0
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