Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.
RBSE Class 10 Maths Solutions Chapter 4 Quadratic Equations 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
Solution:
(i) According to the question
(x + 1)2 = 2(x – 3)
or x2 + 1 + 2x = 2x – 6
or x2 + 1 + 2x – 2x + 6 = 0
or x2 + 7 = 0
or x2 + 0x + 7 = 0
Which is an equation of the form ax2 + bx + c = 0. (a ≠ 0)
∴ It is a quadratic equation.
(ii) According to the question
x2 – 2x = (-2)(3 – x)
or x2 – 2x = -6 + 2x
or x2 – 2x + 6 – 2x = 0
or x2 – 4x + 6 = 0
Which is an equation of the form ax2 + bx + c = 0; (a ≠ 0)
∴ It is a quadratic equation.
(iii) According to the question
(x – 2) (x + 1) = (x – 1) (x + 3)
or x2 + x – 2x – 2 = x2 + 3x – x – 3
or x2 – x – 2 = x2 + 2x – 3
or x2 – x – 2 – x2 – 2x + 3 = 0
or – 3x + 1 = 0
Which contain no terms of x2
∴ It is not a quadratic equation.
(iv) According to the question
(x – 3) (2x + 1) = x (x + 5)
or 2x2 + x – 6x – 3 = x2 + 5x
or 2x2 – 5x – 3 – x2 – 5x = 0
or x2 – 10x – 3 = 0
Which is an equation of the form ax2 + bx + c = 0; (a ≠ 0)
∴ It is a quadratic equation.
(v) According to the question
(2x – 1) (x – 3) = (x + 5) (x – 1)
or 2x2 – 6x – x + 3 = x2 – x + 5x – 5
or 2x2 – 7x + 3 = x2 + 4x – 5
or 2x2 – 7x + 3 – x2 – 4x + 5 = 0
or x2 – 11x + 8 = 0
Which is an equation of the form ax2 + bx + c = 0; (a ≠ 0)
∴ It is a quadratic equation.
(vi) According to the question
x2 + 3x + 1 = (x – 2)2
or x2 + 3x + 1 = x2 + 4 – 4x
or x2 + 3x + 1 – x2 – 4 + 4x = 0
or 7x – 3 = 0
Which is an equation of the form ax2 + bx + c = 0; (a ≠ 0)
∴ It is a quadratic equation.
(vii) According to the question
(x + 2)3 = 2x(x2 – 1)
or x3 + (2)3 + 3(x)2 2 + 3(x)(2)2 = 2x3 – 2x
or x3 + 8 + 6x2 + 12x = 2x3 – 2x
or x3 + 8 + 6x2 + 12x – 2x3 + 2x = 0
or -x3 + 6x2 + 14x + 8 = 0
Here the highest power of x is 3
Therefore it is a cubic equation.
∴ It is not a quadratic equation.
(viii) According to the question x3 – 4x2 – x + 1 = (x – 2)3
or x3 – 4x2 – x + 1 = x3 – (2)3 + 3(x)2(-2) + 3(x) (- 2)2
or x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
or x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0
or 2x2 – 13x + 9 = 0
Which is an equation of the form ax2 + bx + c = 0; (a ≠ 0)
∴ It is a quadratic equation.
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.
Solution:
(i) Let the breadth of the rectangular plot = x m
∴ Length of the rectangular plot = (2x + 1) m
∴ Area of the rectangular plot = [x(2x + 1)]m2
= (2x2 + x) m2
According to the question
2x2 + x = 528
or 2x2 + x – 528 = 0
Hence the quadratic equation is
2x2 + x – 528 = 0
Where x (in metres) is the breadth of the plot.
(ii) Let x and x + 1 be two consecutive positive integers.
Then,
Product of the integers = x (x + 1)
= x2 + x
According to the question
x2 + x = 306
or x2 + x – 306 = 0
Hence the given problem in the form of a quadratic equation is
x2 + x – 306 = 0
where x is the smaller integer.
(iii) Let Rohan’s present age= x years
Then, Age of Rohan’s mother = (x + 26) years
3 years from now,
Rohan’s age = (x + 3) years
and Age of Rohan’s mother = (x + 26 + 3) years
= (x + 29) years
Their product = (x + 3) (x + 29)
= x2 + 29x + 3x + 87
= x2 + 32x + 87
According to the question,
x2 + 32x + 87 = 360
or x2 + 32x + 87 – 360 = 0
or x2 + 32x – 273 = 0
Hence the given problem in the form of a quadratic equation is
x2 + 32x – 273 = 0
where x (in years) is the present age of Rohan.
(iv) Let the speed of the train be u km./h. Then,
time taken in covering a distance of 480 km = \(\frac{480}{u}\) hours [∵ time = \(\frac{Distance}{Speed}\)]
If the speed of the train had been 8 km/ h less, then time taken in covering the same distance = \(\frac{480}{u-8}\)hours
According to the question 3 hours are taken more to cover the same distance
Therefore \(\frac{480}{u-8}\) – \(\frac{480}{u}\) = 3
or \(\frac{160}{u-8}\) – \(\frac{160}{u}\) = 1
or 160u – 160 (u – 8) = u(u – 8)
or 160u – 160u + 1280 = u2 – 8u
or u2 – 8u – 1280 = 0
Hence the quadratic equation form of the given problem is
u2 – 8u – 1280 = 0
Where u (in km/h) is the speed of the train.
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