Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Questions and Answers.
BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 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) We have:
(x + 1)2 = 2(x – 3)
or x2 + 2x + 1 = 2x – 6
or x2 + 2x + 1 – 2x + 6 = 0
or x2 + 7 = 0
Clearly, x2 + 7 is a quadratic polynomial. So, the given equation is a quadratic equation.
(ii) We have:
x2 – 2x = (- 2)(3 – x)
or x2 – 2x + 2(3 – x) = 0
or x2 – 2x + 6 – 2x = 0
or x2 – 4x + 6 = 0
Clearly, x2 – 4x + 6 is a quadratic polynomial. So, the given equation is a quadratic equation.
(iii) We have:
(x – 2)(x + 1) = (x – 1)(x + 3)
or x2 – x – 2 = x2 + 2x – 3
or x2 – x – 2 – x2 – 2x + 3 = 0
or – 3x + 1 = 0
Clearly, – 3x + 1 is a linear polynomial, i.e., it is not a quadratic polynomial. So, the given equation is not a quadratic equation.
(iv) We have:
(x – 3)(2x + 1) = x(x + 5)
or x(2x + 1) – 3(2x + 1) – x(x + 5) = 0
or 2x2 + x – 6x – 3 – x2 – 5x = 0
or x2 – 10x – 3 = 0
Clearly, x2 – 10x – 3 is a quadratic polynomial. So, the given equation ,is a quadratic equation.
(v) We have:
(2x – 1)(x – 3) = (x + 5)(x – 1)
or (2x – 1)(x – 3) – (x + 5)(x – 1) = 0
or 2x(x – 3) – 1(x – 3) – x(x – 1) – 5(x – 1) = 0
or 2x2 – 6x – x + 3 – x2 + x – 5x + 5 = 0
or x2 – 11x + 8 = 0
Clearly, x2 – 11x + 8 is a quadratic polynomial. So, the given equation is a quadratic equation.
(vi) We have:
x2 + 3x + 1 = (x – 2)2 = 0
or x2 + 3x + 1 – (x – 2)2 = 0
or x2 + 3x + 1 – (x2 – 4x + 4) = 0
or x2 + 3x + 1 – x2 + 4x – 4 = 0
or 7x – 3 = 0
Clearly, 7x – 3 is not a quadratic polynomial. So, the given equation is not a quadratic equation.
(vii) We have:
(x + 2)3 = 2x(x2 – 1)
or x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
or x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
or – x3 + 6x2 + 14x + 8 = 0
Clearly, – x3 + 6x2 + 14x + 8, being a polynomial of degree 3, is not a quadratic polynomial.
So, the given equation is not a quadratic equation.
(viii) We have:
x3 – 4x2 – x + 1 = (x – 2)3
= x3 + 3x2(- 2) + 3x(- 2)2 + (- 2)3
= x3 – 6x2 + 12x – 8
or x3 – 4x2 – x + 1 – x3 + 6x2 – 12x + 8 = 0
or 2x2 – 13x + 9 = 0
Clearly, 2x2 – 13x + 9 is a quadratic polynomial. So, the given equation 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 length and breadth of the rectangular plot be 2x + 1 metres and x metres, respectively. It is given that its area = 528 m2.
∴ (2x + 1) × x = 528 or 2x2 + x = 528 or 2x2 + x = 528
or 2x2 + x – 528 = 0,
which is the required quadratic equation satisfying the given conditions.
(ii) Let two consecutive integers be x and x + 1 such that their product = 306.
or x(x + 1) = 306 or x2 + x – 306 = 0,
which is the required quadratic equation satisfying the given conditions.
(iii) Let Rohan’s present age be x years. Then,
his mother’s age = (x + 26) years
After 3 years, their respective ages will be (x + 3) years and (x + 29) years. ………………. (1)
It is given that the product of ages mentioned at (1) will be 360.
i. e., (x + 3)(x + 29) = 360
or x2 + 32x + 87 = 360
or x2 + 32x + 87 – 360 = 0
or x2 + 32x – 273 = 0
Therefore, the age of Rohan satisfies the quadratic equation x2 + 32x – 273 = 0.
(iv) Let u km/h be the speed of the train.
Then, time taken to cover 480 km = \(\frac{480}{u}\) hours.
Time taken to cover 480 km when the speed is decreased by 8 km/h
= \(\frac{480}{u-8}\) hours.
It is given that the time to cover 480 km is increased by 3 hours.
∴ \(\frac{480}{u-8}\) – \(\frac{480}{u}\) = 3
or 480u – 480(u – 8) = 3u(u – 8)
or 160u – 160u + 1280 = u2 – 8u
or u3 – 8u – 1280 = 0.