Class 10 - Mathematics
Quadratic Equations -Exercise 4.1
Top Block 1
Exercise 4.1
Question : 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
Answer :
(i) (x + 1)2 = 2(x – 3)
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 7 = 0
⇒ x2 + 0x + 7 = 0
This is an equation of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.
(ii) x2 – 2x = (–2) (3 – x)
⇒ x2 – 2x = -6 + 2x
⇒ x2 -2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
This is an equation of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x2 – 2x + x – 2 = x2 – x + 3x – 3
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ x2 – x – 2 – x2 – 2x + 3 = 0
⇒ -3x + 1 = 0
⇒ 3x – 1 = 0
This is not an equation of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.
(iv) (x – 3)(2x +1) = x(x + 5)
⇒ 2x2 – 6x + x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x = 0
⇒ x2 – 10x – 3 = 0
This is an equation of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x2 – x – 6x + 3 = x2 + 5x – x – 5
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ 2x2 – 7x + 3 – x2 – 4x + 5 = 0
⇒ x2 – 11x + 8 = 0
This is an equation of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ x2 + 3x + 1 – x2 – 4 + 4x = 0
⇒ 7x – 3 = 0
This is not an equation of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.
(vii) (x + 2)3 = 2x (x2 – 1)
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ -x3 + 6x2 + 14x + 8 = 0
⇒ x3 – 6x2 – 14x – 8 = 0
This is not an equation of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.
(viii) x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ x3 – 4x2 – x + 1 – x3 + 6x2 – 12x + 8 = 0
⇒ 2x2 – 13x + 9 = 0
This is an equation of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.
Mddle block 1
Question : 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.
Answer :
(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
(ii) Let the two consecutive numbers be x and (x + 1).
Given, product of the numbers = 306
⇒ x(x + 1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Thus, the required quadratic equation is: x2 + x – 306 = 0
(iii) Let the present age of Rohan = x
So, his mother’s age = (x + 26) years
After 3 years,
Rohan’s age = (x + 3) years
His mother’s age = [(x + 26) + 3] years = (x + 29) years
According to the condition,
The product of their ages (in years) 3 years from now will be 360
⇒ (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
So, time taken = Distance/speed = 480/(u – 8) hours
According to the condition,
480/(u – 8) – 480/u = 3
⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u2 – 24u
⇒ 3840 – 3u2 + 24u = 0
⇒ 1280 – u2 + 8u = 0
⇒ –1280 + u2 – 8u = 0
⇒ u2 – 8u – 1280 = 0
Thus, the required quadratic equation is: u2 – 8u – 1280 = 0
