Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Solution

 

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 = 480u hours.
Time taken to cover 480 km when the speed is decreased by 8 km/h
= 480u−8 hours.
It is given that the time to cover 480 km is increased by 3 hours.
∴ 480u−8 – 480u = 3
or 480u – 480(u – 8) = 3u(u – 8)
or 160u – 160u + 1280 = u2 – 8u
or u3 – 8u – 1280 = 0.

BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x3 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0
(iv) 2x2 – x + \(\frac{1}{8}\) = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) We have:
x2 – 3x – 10 = 0
or x2 – 5x + 2x – 10 = 0
or x(x – 5) + 2(x – 5) = 0
or (x – 5)(x + 2) = 0
i.e; x – 5 = 0 or x + 2 = 0
i.e; x = 5 or x = – 2
Thus, x = 5 and x = – 2 are two solutions of the equation x2 – 3x – 10 = 0. That is, roots of the equation are 5 and – 2.

(ii) We have:
2x2 + x – 6 = 0
or 2x2 + 4x – 3x – 6 = 0
or 2x(x + 2) – 3(x + 2) = 0
or (x + 2)(2x – 3) = 0
i.e; x + 2 = 0 or 2x – 3 = 0
i.e; x = – 2 or x = \(\frac{3}{2}\)
Thus, – 2 and \(\frac{3}{2}\) are two roots of the equation 2x2 + x – 6 = 0.

(iii) We have:
\(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0
or \(\sqrt{2}\)x(x + \(\sqrt{2}\)) + 5(x + \(\sqrt{2}\)) = 0
or (x + \(\sqrt{2}\))(\(\sqrt{2}\)x + 5) = 0
i.e; x + \(\sqrt{2}\) = 0 or \(\sqrt{2}\)x + 5 = 0
i.e; x = – \(\sqrt{2}\)
or x = \(\frac{-5}{\sqrt{2}}=\frac{-5 \sqrt{2}}{2}\)
Thus, – \(\sqrt{2}\) and \(\frac{-5 \sqrt{2}}{2}\) are two roots of the equation \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0.

(iv) We have:
2x2 – x + \(\frac{1}{8}\) = 0
or 16x2 – 8x + 1 = 0
or 16x2 – 4x – 4x + 1 = 0
or 4x(4x – 1) – 1(4x – 1) = 0
or 4x – 1 = 0 or 4x – 1 = 0
i.e; x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)
Thus, \(\frac{1}{4}\) and \(\frac{1}{4}\) are two equal roots of the equation 2x2 – x + \(\frac{1}{8}\) = 0

(v) We have:
100x2 – 20x + 1 = 0
or 100x2 – 10x – 10x + 1 = 0
or 10x(10x – 1) – 1(10x – 1) = 0
or (10x – 1)(10x – 1) = 0
i.e; 10x – 1 = 0 or 10x – 1 = 0
i.e; x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)
Thus, \(\frac{1}{10}\) and \(\frac{1}{10}\) are two equal roots of the equation 100x2 – 20x = 1 = 0.

Question 2.
Solve the problems given in Example 1, i.e; to solve
(i) x2 – 45x + 324 = 0
(ii) x2 – 55x + 750 = 0
using factorisation method.
Solution:
(i) We have: x2 – 45x + 324 = 0
or x2 – 9x – 36x + 324 = 0
or x(x – 9) – 36(x – 9) = 0
or (x – 9)(x – 36) = 0
i.e; x – 9 = 0 or x – 36 = 0
i.e; x = 9 or x = 36
Thus, 9 and 36 are two roots of the equation x2 – 45x + 324 = 0.

(ii) We have:
x2 – 55x + 750 = 0
or x2 – 30x – 25x + 750 = 0
or x(x – 30) – 25(x – 30) = 0
or (x – 30)(x – 25) = 0
i.e; x – 30 = 0 or x – 25 = 0
i.e; x = 30 or x = 25
Thus, 30 and 25 are two roots of the equation.
x2 – 55x + 750 = 0.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the required numbers be x and 27 – x. Then,
x(27 – x) = 182
or 27x – x2 = 182
or x2 – 27x + 182 = 0
or x2 – 13x – 14x + 182 = 0
or x(x – 13) – 14(x – 13) = 0
or (x – 13)(x – 14) = 0
i.e; x – 13 = 0 or x – 14 = 0
i.e; x = 13 or x = 14
Hence, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and x + 1.
Then, x2 + (x + 1)2 = 365
or x2 + x2 + 2x + 1 = 365
or 2x2 + 2x – 364 = 0
or x2 + x – 182 = 0
or x(x + 14) – 13(x + 14) = 0
or (x + 14)(x – 13) = 0
i.e; x + 14 = 0 or x – 13 = 0
i.e; x = – 14 or x = 13
Since x, being a positive integer, cannot be negative. Therefore, x = 13.
Hence, the two consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right traiangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the right triangle be x cm. Then, its altitude is (x – 7) cm.
It is given that the hypotenuse = 13 cm
So, \(\sqrt{x^{2}+(x-7)^{2}}\) = 13
or x2 + (x2 – 14x + 49) = 169
or 2x2 – 14x – 120 = 0
or x2 – 7x – 60 = 0
or x2 – 12x + 5x – 60 = 0
or x(x – 12) + 5(x – 12) = 0
or (x – 12)(x + 5) = 0
i.e., x – 12 = 0 or x + 5 = 0
i.e., x = 12 or x = – 5
So, x = 12
[Since side of a triangle can never be negative]
∴ Length of the base = 12 cm
and length of the altitude
= (12 – 7) cm
= 5 cm.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day.
If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
Then, cost of production of each article = Rs (2x + 3)
It is given that the total cost of production = Rs 90
∴ x × (2x + 3) = 90
or 2x2 + 3x – 90 = 0
or 2x2 – 12x + 15x – 90 = 0
or 2x(x – 6) + 15(x – 6) = 0
or (x – 6)(2x + 15) = 0
i.e; x – 6 = 0 or 2x + 15 = 0
i.e; x = 6 or x = \(\frac{-15}{2}\)
So, x = 6 [Since the number of articles produced cannot be negative]
Cost of each article = Rs (2 × 6 + 3) = Rs 15
Hence, the number of articles produced are 6 and the cost of each article is Rs 15.

BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist by, the method of completing the square:
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 43–√x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) The equation 2x2 – 7x + 3 is the same as

∴ 2x2 – 7x + 3 = 0

∴ The roots of the given equation are 3 and 12.

(ii) We have:
2x2 + x – 4 = 0

∴ The roots of the given equation are
−1+33√4 and −1−33√4

(iii) We have:
4x2 + 43–√x + 3 = 0
or (2x)2 + 2 × (2x) × 3–√ + (3–√)2 – (3–√)2 + 3 = 0
or (2x + 3)2 – 3 + 3 = 0
or (2x + 3–√)2 = 0
i.e; x = – 3√2 and – 3√2

(iv) We have,
2x2 + x + 4 = 0
or x2 + 12x + 2 = 0

But (x + 14)2 cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.

Question 2.
Find the roots of the quadratic equations given in Q. 1 above by applying the quadratic formula.
Solution:
(i) The given equation is 2x2 – 7x + 3 = 0.
Here, a = 2, b = – 7 and c = 3.
∴ D = b2 – 4ac = (- 7)2 – 4 × 2 × 3 = 49 – 24 = 25 > 0
So, the given equation has real roots given by

∴ The roots of the given equation are 3 and 12

(ii) The given equation is 2x2 + x – 4 = 0
Here, a = 2, b = 1 and c = – 4.
∴ D = b2 – 4ac = (1)2 – 4 × 2 × (- 4) = 1 + 32 = 33 > 0
So, the given equation has real roots given by

∴ The roots of the given equation are
−1+33√4 and −1−33√4

(iii) The given equation is 4x2 + 43–√x + 3 = 0
Here, a = 4, b = 43–√ and c = 3.
∴ D = b2 – 4ac = (43–√)2 – 4 × 4 × 3 = 48 – 48 = 0
So, the given equation has real equal roots given by

∴ The equal roots of the given equation are
−3√2 and −3√2

(iv) The given equation is 2x2 + x + 4 = 0
Here, a = 2, b = 1 and c = 4.
∴ D = b2 – 4ac = (1)2 – 4 × 2 × 4 = 1 – 32 = – 31 < 0
So, the given equation has no real roots.
[Sometimes, it is easier to use quadratic formula method as it is a straight forward method.]

Question 3.
Find the roots of the following equations:
(i) x – 1x = 3, x ≠ 0
(ii) 1x+4 – 1x+7 = 1130, x ≠ – 4, 7
Solution:
(i) The given equation is x – 1x = 3, x ≠ 0.
So, x2 – 3x – 1 = 0
Here, a = 1, b = – 3 and c = – 1.
∴ D = b2 – 4ac = (- 3)2 – 4(1)(- 1) = 9 + 4 = 13 > 0
So, the given equation has real roots given by

∴ The roots of the given equations are
3+13√2 and 3−13√2

(ii) The given equation is 1x+4 – 1x−7 = 1130, x ≠ – 4, 7

or – 30 = x2 – 3x – 28
or x2 – 3x + 2 = 0
or (x – 1)(x – 2) = 0
i.e; x – 1 = 0 or x – 2 = 0
⇒ x = 1 or x = 2
Thus, 1 and 2 are the roots of the given equation.

Question 4.
The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 13. Find his present age.
Solution:
Let Rehman’s present age be x years
As per question, we have:
1x−3 + 1x+5 = 13
or 3(x + 5) + 3(x – 3) = (x – 3)(x + 5)
or 3x + 15 + 3x – 9 = x2 + 2x – 15
or x2 – 4x – 21 = 0
or (x – 7)(x + 3) = 0 i.e., x = 7 or x = – 3
x = 7 [Since age can never be negative]
Thus, Rehman’s present age is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the
product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x. Then, her marks in English will be (30 – x).
As per condition of the problem:
(x + 2) × [(30 – x) – 3] = 210
or (x + 2)(27 – x) = 210
or 27x – x2 + 54 – 2x = 210
or x2 – 25x + 156 = 0
or x2 – 12x – 13x + 156 = 0
or x(x – 12) – 13(x – 12) = 0
or (x – 12)(x – 13) = 0
i.e., x = 12
or x = 13
∴ Shefali’s marks in Mathematics and English are 12 and 18, respectively on her marks in Mathematics and English are 12 and 18, respectively.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let in the rectangular field ABCD, BC = x metres.
Then AC = (x + 60) mctrcs and AB (x + 30) metres.
By Pythagoras Theorem, we have:

AC2 = BC2 + AB2
So, (x + 60)2 = x2 + (x + 30)2
or x2 + 120x + 3600 = x2 + x2 + 60x + 900
or x2 – 60x – 2700 = 0
or x2 – 90x + 30x – 2700 = 0
or x(x – 90) + 30(x – 90) = 0
i.e; x = – 30 or x = 90
∴ x = 90 [∵ Side of a rectangle can never be negative]
Hence, the sides of the field are 120 m and 90 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be x. Then,
Square of the smaller number = 8x
Also, square of the larger number = x2
It is given that the difference of the squares of the numbers is 180.
∴ x2 – 8x = 180
or x2 – 8x – 180 = 0
or x2 – 18x + 10x – 180 = 0
or x(x – 18) + 10(x – 18) = 0
or (x – 18)(x + 10) = 0
i.e; x – 18 = 0 or x + 10 = 0
i.e; x = 18 or x = – 10

Case I:
When x = 18.
In this case, we have:
Square of the smaller number = 8x = 8 × 18 = 144
∴ Smaller number = ± 12
Thus, the numbers are 18, 12 or 18, – 12.

Case II:
When x = – 10
In this case, we have:
Square of the smaller number = 8x = 8 × – 10 = – 80
But, square of a number is always positive.
∴ x = – 10 is not possible.
Hence, the numbers are 18, 12 or 18, – 12.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let x km/h be the uniform speed of the train.
Then, time taken to cover 360 km when the speed is increased by 5 km/hr = 360x hours.
It is given that the time to cover 360 km is reduced by 1 hour.
∴ 360x – 360x+5 = 1
or 360(x + 5) – 360x = x(x + 5)
or 360x + 1800 – 360x = x2 + 5x
or x2 + 5x – 1800 = 0
or x2 + 45x – 40x – 1800 = 0
or x(x + 45) – 40(x + 45) = 0
or (x – 40)(x + 45) = 0
i.e., x – 40 = 0 or x + 45 = 0
i.e., x = 40 or x = – 45
But speed cannot be negative. Therefore, x = 40. Hence, the original speed of the train is 40 km/h.

Question 9.
Two water taps together can fui a tank in 938 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately.
Find the time in which each tap can separately fill the tank.
Solution:
Let the smaller tap takes x hours to fill the tank. So, the larger tap will take (x – 10) hours to fill the tank.
∴ Portion of the tank filled by the larger tap in one hour
= 1x−10
So, Portion of the tank filled by the larger tap in 938 hours, i.e., 758 hours = 1x−10 × 758
Similarly, portion of the tank filled by the smaller tap in 758 hours = 1x × 758
Since it is given that the tank is completely filled in 758 hours, therefore

or 150x – 750 = 8x2 – 80x
or 8x2 – 230x + 750 = 0
or 4x2 – 115x + 375 = 0
or 4x2 – 100x – 15x + 375 = 0
or 4x(x – 25) – 15(x – 25) = 0
or (x – 25)(4x – 15) = 0
i.e; x – 25 = 0 or 4x – 15 = 0
i.e; x = 25 or 154
∴ x = 25, as x = 154 is inadmissible (154 < 10)
Hence, the larger tap fills the tank in 15 hours and the smaller tap takes 25 hours to fill the tank.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations).
If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the speed of the express train be x km/h.
Then, the speed of the passenger train is (x – 11) km/h
Total distance to be covered = 132 km
Time taken by express train = 132x hours
and tame taken by the passenger train = 132x−11 hours
By the given condition, 132x−11 – 132x = 1
or 132x – 132x + 1452 = x2 – 11x
or x2 – 11x – 1452 = 0
or x2 – 44x + 33x – 1452 = 0
or x(x – 44) + 33(x – 44) = 0
or (x – 44)(x + 33) = 0
i.e; x = 44 or x = – 33
But speed cannot be negative. Therefore, x = 44
∴ Speed of express train = 44 km/hr
and speed of passenger train = (44 – 11) km/h
= 33 km/h

Question 11.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the sides of the squares be x and y metres (x > y).
According to question:
x2 + y2 = 468 ……………… (1)
and 4x – 4y = 24
i.e., x – y = 6 …………… (2)
Putting x = y + 6 in (1), we get
(y + 6)2 + y2 = 468
or 2y2 + 12y + 36 – 468 = 0
or y2 + 6y – 216 = 0
or y2 + 18y – 12y – 216 = 0
or y(y + 18) – 12(y + 18) = 0
or (y + 18)(y – 12) = 0
y = – 18 or y = 12
But side cannot be negative. Therefore, y = 12.
∴ x = y + 6 = 12 + 6 = 18
∴ The sides of the squares are 18 m and 12 m.

BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 43–√x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) The given equation is 2x2 – 3x + 5 = 0.
Here, a = 2, b = – 3 and c = 5.
∴ D = b2 – 4ac = (- 3)2 – 4 × 2 × 5 = 9 – 40 = – 31 < 0
So, the given equation has no real roots.

(ii) The given’equation is 3x2 – 43–√x + 4 = 0
Here, a = 3, b = – 43–√ and c = 4.
∴ D = b2 – 4ac = (- 43–√)2 – 4 × 2 × 5 = 9 – 40 = – 31 < 0
So, the given equation has real equal roots, given by

(iii) The given equation is 2x2 – 6x + 3 = 0.
Here, a = 2, b = – 6 and c = 3
∴ D = b2 – 4ac = (- 6)2 – 4 × 2 × 3 = 36 – 24 = 12 > 0
So, the given equation has real roots, given by

∴ The roots of the given equation are 3+3√2 and 3−3√2.

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(i) 2x2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) The given equation is 2x2 + kx + 3 = 0.
Here, a = 2, b = k and c = 3
∴ D = b2 – 4ac = k2 – 4 × 2 × 3 = k2 – 24
The given equation will have real and equal roots, if
D = 0 i.e; k2 – 2kx + 6 = 0
Here, a = k, b = – 2k and c = 6
∴ D = b2 – 4ac = (- 2k)2 – 4 × k × 6 = 4k2 – 24k
The given equation will have real and equal roots, if D = 0.
So, 4k2 – 24k = 0 or 4k(k – 6) = 0
i.e., k = 0 or k = 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:
Let 2x be the length and x be the breadth of a rectangular mango grove.
Area = (2x)(x) = 800. [Given]
So, x2 = 400 or x = 20 [Since side cannot be negative]
The value of x is real so to design of grove is possible
Its length = 40 m and breadth = 20 m.

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let age of one of the friends = x years. Then, age of the other friend = 20 – x.
Four years ago,
Age of one of the friends = (x – 4) years
and age of the other friend = (20 – x – 4) years
= (16 – x) years

According to condition:
(x – 4)(16 -x) = 48
or 16x – x2 – 64 + 4x = 48
or x2 – 20x + 112 = 0
Here, a = 1, b = – 20 and c = 112.
∴ D = b2 – 4ac = (- 20)2 – 4 × 1 × 22
= 400 – 448 = – 48 < 0
So, the given equation has no real roots.
Thus, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Let length be x metres and breadth be y metres.
∴ Perimeter = 80 m
or 2(x + y) = 80 or x + y = 40 ……………….. (1)
Also, Area = 400 m2
So, xy = 400
or x(40 – x) = 400 [Using (1)]
or 40x – x2 = 400 or x2 – 40x + 400 = 0
Here, a = 1, b = – 40 and c = 400.
∴ D = b2 – 4ac = (- 40)2 – 4 × 1 × 400
= 1600 – 1600 = 0
So, the given equation has equal real roots. To design a rectangular park is possible.
∴ Its length and breadth is given by
x2 – 40x + 400 = 0
or (x – 20)2 = 0 i.e., x – 20 = 0 or x = 20
∴ Length = 20 m, Breadth = (40 – 20) m = 20 m.