Class X Math
Quadratic Equations
Quadratic Equations
EXERCISE 4.1
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 – × + 1 = (x – 2)3
Solution. (i) (x + 1)2 = 2(x – 3)
We have:
(x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 70
Since x2 + 7 is a quadratic polynomial
∴ (x + 1)2 = 2(x – 3) is a quadratic equation.
Solution. (ii) x2– 2x = (–2) (3 – x)
We have:
x2 – 2x = (– 2) (3 – x)
⇒ x2 – 2x = –6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
Since x2 – 4x + 6 is a quadratic polynomial
∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.
Solution. (iii) (x – 2) (x + 1) = (x – 1) (x + 3)
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
Since –3x + 1 is a linear polynomial
∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.
Solution. (iv) (x – 3) (2x + 1) = x(x + 5)
We have:
(x – 3) (2x + 1) = x(x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x – 0
⇒ x2 + 10x – 3 = 0
Since x2 + 10x – 3 is a quadratic polynomial
∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
Solution. (v) (2x – 1) (x – 3) = (x + 5) (x – 1)
We have:
(2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
⇒ 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0
⇒ x2 – 11x + 8 = 0
Since x2 – 11x + 8 is a quadratic polynomial
∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation.
Solution. (vi) x2 + 3x + 1 = (x – 2)2
We have:
x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
⇒ x2 + 3x + 1 – x2 + 4x – 4 =0
⇒ 7x – 3 = 0
Since 7x – 3 is a linear polynomial.
∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.
Solution. (vii) (x + 2)3 = 2x(x2 – 1)
We have:
(x + 2)3 = 2x(x2 – 1)
x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ –x3 + 6x2 + 14x + 8 = 0
Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3
∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation.
Solution. (viii) x3 – 4x2 – x + 1 = (x – 2)3
We have:
x3 – 4x2 – x + 1 = (x – 2)3
⇒ 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
Since 2x2 – 13x + 9 is a quadratic polynomial
∴ x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic equation.
Q.2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. 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.
Solution.
(i) Let the breadth = x metres
Length = 2 (Breadth) + 1
Length = (2x + 1) metres
Since Length × Breadth = Area
∴ (2x + 1) × x = 528
2x2 + x = 528
2x2 + x – 528 = 0
Thus, the required quadratic equation is
2x2 + x – 528 = 0
Solution.
(ii) Let the two consecutive numbers be x and (x + 1).
∵ Product of the numbers = 306
∴ x (x + 1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Thus, the required equdratic equation is
x2 + x – 306 = 0
Solution.
(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
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 29x + 3x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Thus, the required quadratic equation is
x2 + 32x – 273 = 0
(iv) Let the speed of the tram = u km/hr
Distance covered = 480 km
Time taken = Distance + Speed
= (480 ÷ u) hours
In second case,
Speed = (u – 8) km/ hour
According to the condition,
⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u2 – 24u
⇒ 3840 – 3u2 + 24u = 0
⇒ 1280 – u2 + 8u = 0
⇒ u2 – 8u – 1280 = 0
Thus, the required quadratic equation is
u2 – 8u – 1280 = 0