Exercise 4.1
Multiple Choice Questions (MCQs)
Question1. Which of the following is a quadratic equation?

Solution: (d)
Explanation:
An equation which is of the form ax² + bx + c = 0, a ≠ 0 is called a quadratic equation.

⟹ x³ – x² = x³ -3x² + 3x – 1
⟹-x² + 3x² – 3x + 1 = 0
⟹ 2x² – 3x + 1 = 0
This equation is not of the form ax²+ bx + c, a ≠ 0. Thus, the equation is a quadratic equation.
Question2. Which of the following is not a quadratic equation?


This equation is not of the form ax² + bx + c, a ≠ 0.
Thus, the equation is not quadratic.
Question3. Which of the following equations has 2 as a root?
(a) x² -4x + 5 = 0       
(b) x² + 3x -12 = 0
(c) 2x² -7x + 6 = 0     
(d) 3x² – 6x -2 = 0
Solution: (c)
Explanation:
If k is one of the root of any quadratic equation then x = k
i.e., k satisfies the equation ak² + bk + c = 0.
i.e., f(k) = ak² + bk + c = 0,
(a) Putting x = 2 in x² – 4x + 5, we get
(2)² – 4(2) + 5 = 4 – 8 + 5 = 1≠0.
So, x = 2 is not a root of x² – 4x + 5 = 0.
(b) Putting x = 2 in x² + 3x – 12, we get
(2)² + 3(2) – 12 = 4 + 6 – 12 = –2 ≠ 0
So, x = 2 is not a root of x² + 3x – 12 ≠ 0.
(c) Putting x = 2 in 2x² – 7x + 6, we get
2(2)² – 7(2) + 6 = 2 (4) – 14+ 6 = 8 – 14 + 6 = 14 – 14 = 0
So, x = 2 is root of the equation 2x² – 7x + 6 = 0.
(d) Putting x = 2 in 3x² –6x –2, we get
3(2)² –6(2) –2 = 12 –12 –2 = –2 ≠ 0
So, x = 2 is not a root of 3x² – 6x –2 = 0.

⟹ k = 2
Question5. Which of the following equations has the sum of its roots as 3?


So, option (b) is correct.
Question6. Value(s) of k for which the quadratic equation 2x² –kx + k = 0 has equal roots is/are
(a) 0  
(b) 4  
(c) 8  
(d) 0, 8
Solution: (d)
Explanation:
For a quadratic equation of the form, ax² + bx + c = 0, a ≠ 0 to have two equal real roots, its discriminant must be equal to zero, i.e., D = b² –4ac = 0
Given equation is 2x² – kx + k = 0
On comparing with ax² + bx + c = 0, we get
a = 2, b = – k and c = k
For equal roots, the discriminant must be zero.
i.e., D = b² – 4ac = 0
⟹ (– k)² – 4 (2) k = 0
⟹ k² – 8k = 0
⟹ k (k – 8) = 0
Thus, k = 0, 8
Hence, the required values of k are 0 and 8.





Question10. Which of the following equations has no real roots?

Hence, the equation has two distinct real roots.

Question11. (x² + l)² – x² = 0 has
(a) four real roots       
(b) two real roots        
(c) no real roots          
(d) one real root
Solution: (c)

Exercise 4.2
Very Short Answer Type Questions
Question11. State whether the following quadratic equations have two distinct real roots. Justify your answer.

Solution:
A quadratic equation, ax² + bx + c = 0; a ≠ 0 will have two distinct real roots if its discriminant, D = b² – 4ac > 0.
(i) Given equation is x² – 3x + 4 = 0
On comparing with standard equation we have,
a = 1, b = –3 and c = 4
Now, D = b² – 4ac = (–3)² –4(1) (4) = 9 – 16 = – 7< 0
Hence, the equation x² –3x + 4 = 0 has no real roots.
(ii) Given equation is, 2x² + x – 1 = 0
On comparing with standard equation we have,
a = 2, b = 1 and c = –1
Now, D= b² – 4ac = (1)² – 4(2) (–1) = 1 + 8 = 9 > 0
Hence, the equation 2x² + x – 1 = 0 has two distinct real roots.

(iv) Given, 3x² –4x + 1 = 0
On comparing with standard equation we have,
a = 3, b = –4 and c = 1
Now, D= b² –4ac = (–4)² – 4(3) (1) = 16 – 12 = 4 > 0
Hence, the equation 3x² – 4x + 1 = 0 has two distinct real roots.
(v) Given, (x + 4)² –8x = 0
Simplifying the above equation, we have:
            x²+ 16 + 8x – 8x = 0        [∵(a + b)² = a²+ 2ab + b²]
⟹        x² + 16 = 0
⟹       x² + 0. x + 16 = 0
On comparing with standard equation we have:
a = 1, b = 0 and c = 16
Now, D= b² –4ac = (0)² – 4 (1) (16) = – 64 < 0
Hence, the equation (x + 4)² – 8x = 0 has imaginary roots.

(viii) Given, x (1 – x) -2 = 0
Simplifying above equation we have:
⟹ x – x² − 2 = 0
⟹ x² – x + 2 = 0
On comparing with standard equation we have,
a = 1, b = –1 and c = 2
Now, D= b² – 4ac = (-1)² –4(1) (2)=1 – 8 = – 7 < 0
Hence, the equation x (1 – x) –2 = 0 has imaginary roots.
(ix) Given, (x – 1) (x + 2) + 2 = 0
Simplifying above equation we have:
⟹ x² + x – 2 + 2 = 0
⟹ x² + x + 0 = 0
On comparing with standard equation we have,
a = 1, b = 1 and c = 0
Now, D= b² – 4ac = 1 – 4(1) (0) = 1 > 0
Hence, equation (x – 1) (x + 2) + 2 = 0 has two distinct real roots.
(x) Given, (x + 1) (x – 2) + x = 0
Simplifying above equation we have:
⟹ x² + x – 2x – 2 + x = 0
⟹ x² – 2 = 0
⟹ x² + 0 x – 2 = 0
On comparing with ax² + bx + c = 0, we have
a = 1, b = 0 and c = –2
Now, D= b² –4ac = (0)² – 4(1) (–2) = 0 + 8 = 8 > 0
Hence, the equation (x + 1) (x – 2) + x = 0 has two distinct real roots.
Question 2. Write whether the following statements are true or false. Justify your answers.
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has atleast one real root.
(iii) Every quadratic equation has atleast two roots.
(iv) Every quadratic equation has almost two roots.
(v) If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
(vi) If the coefficient of x²and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.
Solution:
(i) False. For example, a quadratic equation x² – 9 = 0 has two distinct roots – 3 and 3.
(ii) False. For example, equation x² + 4 = 0 has no real root.
(iii) False. For example, a quadratic equation x² – 4x + 4 = 0 has only one root which is 2.
(iv) True, because every quadratic polynomial has almost two roots.
(v) True, because in this case discriminant is always positive. For example, in ax²+ bx + c = 0, as a and c have opposite sign so, ac < 0 ⟹ Discriminant = b² – 4ac > 0.
(vi) True, because in this case discriminant is always negative. For example, in ax²+ bx + c = 0, as b = 0, and a and c have same sign then ac > 0 ⟹ discriminant = b² – 4ac = – 4 a c < 0
Question3. A quadratic equation with integral coefficient has integral roots. Justify your answer.
Solution:
No, a quadratic equcation with integral coefficient (0, ± 1, ± 2, ± 3…) can have fractional or non integral roots.
For example, the quadratic equation 8x² – 2x – 1 = 0 have integral coefficients (8, −2 and −1)

Question4. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Solution:
Yes, a quadratic equation having coefficients as rational number, has irrational roots.
For example, the quadratic equation x² – 4x – 3 = 0 has rational coefficients.
Now, b² – 4ac = 16 + 12 = 28.

Question5. Does there exist a quadratic equation whose coefficient are all distinct irrationals but both the roots are rationals? Why?
Solution:
Yes, there may be a quadratic equation whose coefficients are all distinct irrationals, but both the roots are rational.

Hence roots are rational.
Question6. Is 0.2 a root of the equation x² –0.4 = 0? Justify your answer.
Solution:
No, because 0.2 does not satisfy the given quadratic equation, as,
(0.2)² – 0.4 = 0.04 – 0.4 = − 0.36 ≠ 0.
Question7. If b = 0, c < 0, is it true that the roots of x² + bx + c = 0 are numerically equal and opposite in sign? Justify your answer.
Solution:
Given quadratic equation is, x² + bx + c = 0   …(i)
Also given that b = 0 and c < 0
Putting b = 0 in equation (i), we get:
Hence, roots of x² + bx +c = 0 are numerically equal and opposite in sign.
Exercise 4.3
Short Answer Type Questions
Question1. Find the roots of the quadratic equations by using the quadratic formula in each of the following

Solution:
Concept used: For a quadratic equation, ax² + bx+ c = 0, the roots are represented as:






Q. 2 Find the roots of the following quadratic equations by the factorisation method.

Solution.
Concept used: For a quadratic equation, ax² + bx + c = 0, if any of its coefficient is in fractional form then we first make it integral by suitable method. After that we can find the roots of newly formed quadratic equation by factorisation method, i.e., by splitting the middle term.




Exercise 4.4
Long Answer Type Questions
Question1. Find whether the following equations have real roots. If real roots exist, find them

Solution:
Concept used: For a quadratic equation of the form, ax² + bx + c = 0 to have real roots, its discriminant, D = b² – 4ac > 0.





Question2. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Solution:
Let the required natural number be x.
According to the question,
           x² – 84 = 3 (x + 8)
⟹       x² – 84 = 3x + 24
⟹       x² – 3x – 108 = 0
⟹       x² – 12x + 9x – 108 = 0  [By splitting the middle trem]
⟹       x (x – 12) + 9 (x – 12) = 0
⟹       (x – 12) (x + 9) = 0
⟹       x = 12, ‒ 9
Reject x = ‒ 9 value as a natural number can’t be negative.
Hence, the required natural number is 12.
Question3. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Solution:
Let the natural number be x.

Reject x = – 20 value as a natural number can’t be negative.
Hence, the required natural number is 8.
Exercise 4.4
Long Answer Type Questions (Q. No. 4-8)
Question4. A train, travelling at a uniform speed for 360 km, would have taken 48 min less to travel the same distance, if its speed were 5 km/h more. Find the original speed of the train.
Solution:
Let the original speed of the train = x km/h
Distance covered = 360 km
Time taken by train with original speed,


Ignoring x = – 50, as speed cannot be negative, we have x = 45.
Hence, the original speed of the train = 45 km/h.
Question5. If Zeba were younger by 5 year than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age, what is her age now?
Solution:
Let Zeba’s present age be = x years
∴ Zeba’s age when she was 5 year younger = (x – 5) years
According to the question,
⟹ (x – 5)² = 5x + 11
⟹  x² + 25 – 10x = 5x + 11
⟹  x² – 15x + 14 = 0
⟹  x² – 14x – x + 14 = 0 [By splitting middle term]
⟹  x (x – 14) –1 (x – 14) = 0
⟹  (x – 1) (x – 14) = 0
⟹  (x – 1) = 0 or (x – 14) = 0
⟹  x = 14 or 1
Ignore x = 1 because then her age when she was 5 year younger would be, x – 5 = – 4, which is unacceptable.
Hence, Zeba’s present age is 14 year.
Question6. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Solution:
Let, Nisha’s present age be = x year
Therefore, according to the first condition, Asha’s present age = x² + 2
Nisha grows to her mother’s present age after [(x² + 2) – x] years.
Then, Asha’s age will become (x² + 2) +  [(x² + 2) –x] year.
According to the question,
    (x² + 2) + [(x² + 2) – x] = 10x – 1
⟹  2x² – x + 4 = 10x – 1
⟹  2x² – 11x + 5 = 0
⟹  2x² – 10x – x + 5= 0
⟹  2x (x – 5) –1(x – 5) = 0
⟹  (x – 5) (2x – 1) = 0
⟹  (x – 5) = 0 or (2x – 1) = 0
⟹  x = 5 or 1/2

Hence, present age of Nisha = 5year
And present age of Asha = x² + 2 = (5)² + 2 = 25 + 2 = 27 year.
Question7. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be 1184 m² [see figure]. Find the length and breadth of the pond.

Solution:
Given,  length of the rectangular lawn = L = 50 m
And breadth of the rectangular lawn = B = 40 m
Suppose, the width of area of the grass surrounding the pond be x meter.

Then, length of rectangular pond = l = 50 – (x + x) = 50 – 2x
And breadth of rectangular pond = b = 40 – (x + x) = 40 – 2x
Now, Area of grass surrounding the pond =L × B – l × b = 1184 m²
⟹ 50 × 40 – (50 – 2x) (40 – 2x) = 1184
⟹ 2000 – (2000 – 80x – 100x + 4x²) = 1184
⟹ 80x + 100x – 4x² = 1184  
⟹ x² – 45x + 296 = 0
⟹ x² – 37x – 8x + 296 = 0  [By splitting middle term]
⟹ x (x – 37) – 8 (x – 37) = 0
⟹ (x – 8) (x – 37) = 0
⟹ x = 8 or 37
If x = 37
Then length of pond =50 – 2x =  –24
And breadth of pond = 40 – 2x = –34
But length and breadth cannot be negative, so ignore x = 37.
Hence, with x = 8
Length of pond = 50 – 2x = 50 – 16 = 34 m
And breadth of pond = 40 – 2x = 40 – 16 = 24 m.
Question8. At t minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 min less than t²/4 min. Find t.
Solution:
Total time taken by minute hand to run from 2 pm to 3 pm = 60 minutes.
According to question,

As time can not have negative value,