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Given,

x² – 4kx + k = 0

It’s of the form of ax² + bx + c = 0

Where, a =1, b = -4k, c = k

For the given quadratic equation to have real roots D = b² – 4ac ≥ 0

D = (-4k)² – 4(1)(k) ≥ 0

⇒ 16k² – 4k ≥ 0

⇒ 4k(4k – 1) ≥ 0

⇒ k ≥ 0 and k ≥ \(\frac{1}{4}\)

⇒ k ≥ \(\frac{1}{4}\)

The value of k should be greater than or equal to \(\frac{1}{4}\)to have real roots.

i. The given equation is 

x + 1/x = -2 

∴ x² + 1 = -2x …[Multiplying both sides by x] 

∴ x² + 2x+ 1 = 0 

Here, x is the only variable and maximum index of the variable is 2. 

a = 1, b = 2, c = 1 are real numbers and a ≠ 0. 

∴ The given equation is a quadratic equation.

ii. The given equation is 

(m + 2) (m – 5) = 0

∴ m(m – 5) + 2(m – 5) = 0 

∴ m² – 5m + 2m – 10 = 0 

∴ m² – 3m – 10 = 0

Here, m is the only variable and maximum index of the variable is 2. 

a = 1, b = -3, c = -10 are real numbers and a ≠ 0. 

∴ The given equation is a quadratic equation.

vi. The given equation is m³ + 3m² – 2 = 3m³ 

∴ 3m³ – m³ – 3m² + 2 = 0 

∴ 2m³ – 3m² + 2 = 0 

Here, m is the only variable and maximum index of the variable is not 2. 

∴ The given equation is not a quadratic equation.

i. x² – 7x + 5 = 0 

Comparing the above equation with ax² + bx + c = 0, we get 

a = 1, b = -7, c = 5 

ii. 2m² = 5m – 5 

∴ 2m – 5m + 5 = 0 Comparing the above equation with am² + bm + c = 0, we get 

a = 2, b = -5, c = 5

iii. y² = 7y 

∴ y² – 7y + 0 = 0 

Comparing the above equation with ay² + by + c = 0, we get 

a = 1, b = -7, c = 0

i. The given equation is x² + 5x – 2 = 0

Here, x is the only variable and maximum index of the variable is 2. 

a = 1, b = 5, c = -2 are real numbers and a ≠ 0. 

∴ The given equation is a quadratic equation.

ii. The given equation is 

y² = 5y – 10 

∴ y² – 5y + 10 = 0 

Here, y is the only variable and maximum index of the variable is 2. 

a = 1, b = -5, c = 10 are real numbers and a ≠ 0. 

∴ The given equation is a quadratic equation.

iii. The given equation is 

y² + 1/y = 2 

∴ y³ + 1 = 2y …[Multiplying both sides by y ]

∴ y³ – 2y + 1 = 0 Here, 

y is the only variable and maximum index of the variable is not 2. 

∴ The given equation is not a quadratic equation.

Given: Number of diagonals of a polygon with n-sides = \(\frac{n(n-3)}{2}\)

No. of diagonals of a given polygon = 65 i.e., \(\frac{n(n-3)}{2}\) = 65 

where n is number of sides of the polygon 

⇒ n² – 3n = 2 × 65 

⇒ n² – 3n – 130 = 0 

⇒ n² – 13n + 10n – 130 = 0 

⇒ n(n – 13) + 10(n – 13) = 0 

⇒ (n – 13) (n + 10) = 0 

⇒ n – 13 = 0 (or) n + 10 = 0 

⇒ n = 13 (or) n = -10 

But n can’t be negative. 

∴ n = 13 (i.e.) number of sides = 13.

Also to check 50 as the number of diagonals of a polygon 

∴ \(\frac{n(n-3)}{2}\) = 50 

⇒ n² – 3n = 100 

⇒ n² – 3n – 100 = 0

There is no real value of n for which the above equation is satisfied. 

∴ There can’t be a polygon with 50 diagonals.

Given: 2x² – 3x + 5 = 0 

a = 2; b = -3; c = 5

Discriminant = b² – 4ac 

b² – 4ac = (-3)² – 4(2)(5) 

= 9 – 40 

= -31 < 0 

∴ Roots are imaginary.

Given that the equation has imaginary roots, hence the discriminant is less than 0. 

= 25-4k<0 

When we submit 7 in k the condition above will be satisfied and when we replace 6 the condition will be false. So the least value of k is 7.

Correct option is (C) 2

Standard form of quadratic equation is \(ax^2+bx+c=0\) which is an equation of degree 2.

Correct option is C) 2

For the complex roots it will exists in pair. 

Hence the roots are 1+i and 1-i 

Formula for quadratic equation is (x-a) (x-b) = 0

(x-1-i) (x-1+i) = 0 

x² - x + ix - x + 1 - i - ix + i - i² = 0 

x² - 2x + 2 = 0.

Correct option is B) 0

Correct option is C) Not real

Given: In a class test, the sum of the marks obtained by P in Mathematics and science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180.

To find: his marks in the two subjects.

Solution: Let the marks obtained in Mathematics by P be ‘a’

Given, sum of the marks obtained by P in Mathematics and science is 28.

⇒ Marks obtained in science = 28 – a

Also, if he got 3 marks more in Mathematics and 4 marks less in Science, product of his marks, would have been 180.

⇒ (a + 3)(28 – a – 4) = 180

⇒ (a + 3)(24 – a ) = 180

⇒ 24a - a² + 72 - 21a = 180

⇒ -a² + 21a + 72 = 180

⇒ -a² + 21a + 72 - 180 = 0

⇒ -a² + 21a - 108 = 0

⇒ a² – 21a + 108 = 0

Factorise the above quadratic equation by splitting the middle term:

⇒ a² – 12a – 9a + 108 = 0

⇒ a(a – 12) – 9(a – 12) = 0

⇒ (a – 9)(a – 12) = 0

⇒ a = 9 or 12If marks obtained in mathematics is 9, the marks obtained in science is 28-a = 28 - 9 = 19

⇒ Marks in Mathematics = 9, Marks in Science = 19

Or

If marks obtained in mathematics is 12, the marks obtained in science is 28-a = 28 - 12 = 16

Marks in Mathematics = 12, Marks in Science = 16

Let the length of the shorter side b x cm. Then, the length of the longer side = (x + 5) cm.

Since the triangle is right-angled, the sum of the squares of the sides must be equal to the square of the hypotenuse (Pythagoras Theorem).

x² + (x + 5)² = 235²

or x² + x² + 10x + 25 = 625 

or 2x² + 10x - 600 = 0

or x² + 5x - 300 = 0 

or (x + 20) (x - 15) = 0

This gives x = 15 or x = - 20 

We reject x = - 20 and take x = 15.

Thus, length of shorter side = 15 cm. 

Length of longer side = (15 + 5) cm, i.e., 20 cm.

Let the distance of pole from gate A be ‘a’.

Difference of the distance of the pole from two diametrically opposite fixed gates A and B on the boundary is 7 metres.

Distance of pole from gate B = a – 7 m

Diameter of the park = 13 m

Hypotenuse² = length² + breadth²

⇒ 132 = a² + (a – 7)² 

⇒ 169 = 2a² + 49 – 14a 

⇒ a² – 7a – 60 = 0 

⇒ a² – 12a + 5a – 60 = 0 

⇒ a(a – 12) + 5(a – 12) = 0 

⇒ (a + 5)(a – 12) = 0 

⇒ a = 12 m 

Thus distance of pole is 12 m from gate A and 5 m from gate B

Given: The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm.

To find: the lengths of these sides.

Solution: We know(Hypotenuse)² = (perpendicular)² + (base)² ..... (1)

Given, hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm

Let the base be ‘b’.

⇒ Perpendicular = b – 5Put the known values in (1),

⇒ 252 = b² + (b – 5)²

Apply the formula (x – y)² = x² + y² - 2xy in (b – 5)².

Here x = b and y = 5

⇒ 625 = b² + b² + 25 – 10b

⇒ 625 = 2b²+ 25 – 10b

⇒ 2b²+ 25 – 10b = 625

⇒ 2b2+ 25 – 10b - 625 = 0

⇒ 2b²– 10b - 600 = 0

Take out 2 common of the above equation.

⇒ b² – 5b – 300 = 0

Factorise the above quadratic equation by splitting the middle term. 

⇒ b² – 20b + 15b – 300 = 0 

⇒ b(b – 20) + 15(b – 20) = 0

⇒ (b - 20 ) ( b + 15 ) = 0

⇒ (b - 20 ) = 0 and ( b + 15 ) = 0 

⇒ b = 20 and b=-15

Since the length of any side cannot be negative,

We will ignore -15. ⇒ b = 20 

Perpendicular = 20 – 5 = 15 

Hence,Sides are 15 and 20

Let the smaller leg be ‘a’ and longer leg be ‘b’.

Hypotenuse² = length² + breadth²

Given, hypotenuse of a right triangle is \(3\sqrt{10}\) cm

⇒ 9 × 10 = a² + b² 

⇒ a² + b² = 90 -------- (1)

Now, the smaller leg is tripled and the longer leg doubled, new hypotenuse is \(9\sqrt{5}\) cm.

⇒ (3a)² + (2b)² = 81 × 5

⇒ 9a² + 4b² = 405 -------- (2)

Multiplying (1) by 4 and subtracting from eq 2

⇒ 5a² = 45 

⇒ a² = 9 

⇒ a = 3 

Thus, 9 + b² = 90 

⇒ b² = 81 

⇒ b = 9

Let there be x rows. 

Then, the number of students in each row will also be x.

∴ Total number of students = (x² + 24)

According to the question:

(x + 1)² - 25 = x² + 24

⇒ x² + 2x + 1 = 25 - x² - 24 = 0

⇒ 2x - 48 = 0

⇒ 2x = 48

⇒ x = 24

∴ Total number of students = 24² + 24 = 576 + 24 = 600

Area of a square = s² 

Perimeter of a square = 4s 

Let the sides of the square be a and b respectively. 

Given, sum of the areas of two squares is 400 cm² and the difference of their perimeters is 16 cm. 

⇒ a² + b² = 400 and 4a – 4b = 16 

⇒ a – b = 4 

⇒ a = 4 + b 

⇒ (4 + b)² + b² = 400 

⇒ 2b² + 8b + 16 = 400 

⇒ b² + 4b – 192 = 0 

⇒ b² + 16b – 12b – 192 = 0 

⇒ (b + 16)(b – 12 ) = 0 

⇒ b = 12 m 

Thus, a = 16 m

Let the side of the larger square be x.
Let the side of the smaller square be y.
APQ x²+y² = 468
Cond. II 

4x-4y = 24
=> x – y = 6

=> x = 6 + y
x² + y² = 468
=> (6+y)² +y² = 468
on solving we get y = 12
=> x = (12+6) = 18 m
Therefore, sides are 18m & 12m.

Let the cost price be assumed as Rs x.

Given, the dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article.

It’s given that he gains as much as the cost price of the article, thus, Gain% = x%

Gain% = \(\frac{SP - CP}{CP} \times 100\)

x = \(\frac{24 - x}{x} \times 100\)

⇒ x² = (24-x) x 100

⇒ x² + 100x – 2400 = 0

⇒ x² + 120x – 20x – 2400 = 0

⇒ (x + 120)(x – 20) = 0

⇒ x = 20 or -120

Since money cannot be negative, we call neglect -120

⇒ x = 20

Therefore, the cost price of the article is Rs 20.

Time = distance/speed

Now, express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. The average speed of the express trains is 11 km / hr more than that of the passenger train.

Let the average speed of passenger train be ‘a’.

\(\frac{132}{a}-\frac{132}{a+11}=1\)

⇒ 132 ×(a + 11 –a) = a² + 11a 

⇒ a² + 11a – 1452 = 0 

⇒ a² + 44a – 33a – 1452 = 0 

⇒ a(a + 44) – 33(a + 44) = 0 

⇒ (a + 44)(a – 33) = 0 

⇒ a = 33 km/hr 

Speed of express train = 44 km/hr

Solution:

Length = l

Breadth = b

Perimeter = 2(L+b) = 80 

=> l+b = 40         .......... (eq 1)

Area = lb = 400 ........... (eq 2)

=> (l+b)² = (l-b)² + 4lb

=> 40² = (l-b)² + 4 x 400               [From eqn 1 & 2)

=> 1600 = (l-b)² + 1600

=> l-b = 0            .................... (eq 3)

                                L+b = 40

                                L-b =    0

                               2L = 40 => L = 20.

If L=20, then B=20 

So, yes ! We can construct a rectangular park of this length which is a square. Square is a Rectangle. 

Let the age of one of the friends be ‘x’. 

Age of another friend is (20 – x). 

4 years back age of 1st friend is (x – 4) 

4 years back age of 2nd friend = (20 – x – 4) 

= (16 – x) 

Product of their ages is 48. 

∴ (x – 4) (16 – x) = 48 

16x – x² – 64 + 4x = 48 

-x² + 20x – 64 = 48 

-x² + 20x – 64 – 48 = 0 

-x² + 20x – 112 = 0 

x² – 20x + 112 = 0 

Here, a = 1, b = -20, c = 112 

b² – 4ac = (-20)² – 4( 1)( 112) = 400 – 448 = – 48 

Here, b² – 4ac = -48 < 0. 

∴ It has no real roots. 

∴ This situation is not possible.

Solution :- Let present age of one friend be x and that of the other 20 – x.

4 years ago age of first friend = x – 4, and that of the other = 20 – x – 4 = 16 – x.

According to question,  (x – 4 )(16 – x ) = 48

Or, 16x – x² – 64 + 4x =48

Or, x² + 20x – 112 = 0

Or, x² – 20x + 112 = 0

Here, a = 1, b = – 20, c = 112

D = b² – 4ac = (– 20 )² – 4 ×1 × 112 = – 48 < 0.

Hence, the given situation is not possible.

Let’s assume the present age of Shikha be x years

So, 8 years later, age of her = (x + 8) years

Five years ago, her age = (x – 5) years

Given relationship between the ages can be expressed as:

(x – 5)(x +8) = 30

x² + 8x – 5x – 40 = 30

x² + 3x – 40 – 30 = 0

x² + 3x – 70 = 0 [By factorisation method]

x(x – 7) + 10(x – 7) = 0

(x – 7)(x + 10) = 0

x = 7 or x = -10 (neglected)

Since, the age can never be negative.

Therefore, the present age of Shikha is 7 years.

Given: Perimeter of a rectangle 2(1 + b) = 80 

⇒ 6 + b = 80/2 = 40 

Area of the rectangle, l × b = 400 

If possible, let us suppose that length of the rectangle = x m say 

Then its breadth by equation (1) = 40 – x 

By problem area = x . (40 – x) = 400

⇒ 40x – x² = 400 

⇒ x² – 40x + 400 = 0 

Here a = 1; b = -40; c = +400 

b² – 4ac = (-40)² – 4(1)(+400) 

= 1600 – 1600 = 0 

∴ The roots are real and equal. 

They are \(\frac{-b}{2a}\), \(\frac{-b}{2a}\) 

i.e., \(\frac{-(-40)}{2\times1}\) = \(\frac{40}{2a}\) = 20 

∴ The dimensions are 20 m, 20 m. 

(∴ The park is in square shape)

Let the present age of friend 1 be a x years
Given that,
Sum of the ages of two friends = 20 years

⇒ Present age of friend 2 = (20 – x) years 

And also given that, four years ago, the product of their age was 48.
⇒ Age of friend 1 before 4 years = (x – 4) years
And age of friend 2 before 4 years = (20 – x – 4) years = (16 – x) years
Given that,

( − 4)(16 − ) = 48
⇒ 16 − ² − 64 + 4 = 48
⇒ ² − 20 + 112 = 0

Let D be the discriminant of this quadratic equation.
Then, D = (−20)² − 4 × 112 × 1 = 400 − 448 = −48<0
We know that, to have real roots for a quadratic equation that discriminant D must be
greater than or equal to 0 i.e. D ≥ 0

But D< 0 in the above. So, above equation does not have real roots
Hence, the given situation is not possible.

Let the age one of the two friends be x years. 

Then the age of the other = 20 – x 

Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x 

∴ Product of their ages 4 years ago = (x – 4) (16 – x) 

By problem (x – 4) (16 – x) = 48 

⇒ x(16 – x) – 4(16 – x) = 48 

⇒ 16x – x² – 64 + 4x = 48 

⇒ x² – 20x + 112 = 0 

Here a = 1; b = -20; c = 112

b² – 4ac = (-20)² – 4(1) (112) 

= 400 – 448 

= -48 < 0 

Thus the roots are not real. 

∴ The situation is not possible.

Given, Ashu is x years old while his mother Mrs. Veena is x² years old.

After 5 years, Mrs. Veena will be three times old as Ashu.

⇒ x² + 5 = 3(x + 5) 

⇒ x² – 3x – 10 = 0 

⇒ x² – 5x + 2x – 10 = 0 

⇒ x(x – 5) + 2(x – 5) = 0 

⇒ (x + 2)(x – 5) = 0 

⇒ x = 5 km/hr

Given, sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48 

Let the age of one of the friends be ‘a’ 

Age of the other friend = 20 – a 

⇒ (a – 4)(20 –a – 4) = 48 

⇒ (a – 4)(16 – a) = 48 

⇒ a² -20a + 64 + 48 = 0 

⇒ a² – 20a + 112 = 0 D = b² – 4ac 

⇒ D = 400 – 4 × 112 = -48 

Thus, roots are not real as D < 0 

The following situation is not possible

Let the present ages of the younger sister be ‘a’. 

Given, girl is twice as old as her sister. 

Age of elder sister = 2a 

Also, four years ago, the product of their ages (in years) will be 160. 

⇒ (a + 4)(2a + 4) = 160 

⇒ 2a² + 12a + 16 – 160 = 0 

⇒ a² + 6a – 72 = 0 

⇒ a² + 12a – 6a – 72 = 0 

⇒ a(a + 12) – 6(a + 12) = 0 

⇒ (a – 6)(a + 12) = 0 

⇒ a = 6 years 

Age of sisters – 6 years and 12 years

i) True (size of operator) 

ii) False( conditional operator can have 3 operands 

iii) True 

iv) False 

v) False(Multiple assignment is possible,

eg: a=b=c=…..= 5 

vi) True

This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration js as follows, const int bp=100;

(b) -1

Let the roots of the equation x³ – ax² + bx – c = 0 be 

(α – 1), α, (α + 1) 

∴ S₂ = (α – 1)α + α(α + 1) + (α + 1) (α – 1) = b 

⇒ α² – α + α² + α + α² – 1 = b ⇒ 3α² – 1 = b

∴ Minimum value of b = – 1, when α = 0.

(c) a, b

The given equation is (x – a) (x – b) = c 

⇒ x² – (a + b)x + (ab – c) = 0 

As α, β are the roots of this equation, so 

α + β = a + b and αβ = ab – c 

Let γ , δ be the roots of the equation (x – α) (x – β) + c = 0 

i.e., γ , δ are the roots of the equation x² – (α + β)x + (αβ + c)= 0

∴ γ + δ = α + β = a + b                                 ...(i) 

γδ = αβ + c = – c + c = ab                           ...(ii)

∴ From (i) and (ii) we can infer that the roots of the equation 

(x – α) (x – β) + c = 0 are a and b.

(b) 5

log₄ (x – 1) = log₂ (x – 3) 

⇒ log_2² (x – 1) log₂ (x – 3)

⇒ \(\frac{1}{2}\) log₂ (x – 1) log₂ (x – 3)       \(\bigg[\)Using log_mⁿ (x) = \(\frac{1}{n}\) log_mx\(\bigg]\)

⇒ log₂ (x – 1) = 2 log₂ (x – 3) 

⇒ log₂ (x – 1) = log₂ (x – 3)² 

⇒ (x – 1) = (x – 3)²  ⇒ (x – 1) = x² – 6x + 9 

⇒ x² – 7x + 10 = 0 ⇒ (x – 2) (x – 5) = 0 ⇒ x = 2 or 5. 

x = 2 is inadmissible as log₂ (x – 3) is not defined when x = 2.

∴ x = 5.

Put x = –3 in the equation.

(–3)² + 6(–3) + 9

9 – 18 + 9

= 0

Hence it is a solution

Correct option is (B) 65

Given quadratic equation is

\(2x^2+x-8=0\)

\(\therefore\) Discriminant \(=1^2-4\times2\times-8\)

= 1+64 = 65

Correct option is B) 65

(c) no real roots

The discriminant value of a quadratic equation ax² + bx + c = 0, a ≠ 0 is given by,

D = b² - 4ac = 0

i. If D = b² - 4ac > 0, then its roots are distinct and real.

ii. If D = b² - 4ac = 0, then its roots are real and equal.

iii. If D = b² - 4ac < 0, then its roots are imaginary.

Given, 2x² – √5x + 1 = 0

∴ D = b² - 4ac = 0

= (-√5)² - 4(2)(1)

= -3 < 0

Hence, the roots of the quadratic equation 2x² – √5x + 1 = 0 are imaginary.

(a) no real roots

Given, e^{sin x} – e^{–sin x} – 4 = 0 

Let e^{sin x} = y. Then, the given equation becomes 

y – \(\frac{1}{4}\) = 4 ⇒ y² – 4 –1 = 0

⇒ y = \(\frac{4±\sqrt{16+4}}{2}\) = 2 ± √5

⇒ e^{sin x} = 2 ± √5 ⇒ sin x = log_e (2 ± √5)

⇒ sin x = log_e (2 + √5)

(∴ (2 - √5) 0 < and so log_e (2 + √5) − is not defined)

Now (2 + √5) > 4 ⇒ log_e (2 - √5) > 1 

 But the value of sin x lies between –1 and 1, both values inclusive, so sin \(x\) ≠ loge (2 + √5)

∴ There are no possible real roots of the given equation.

(b) 138.

α + β = 6, αβ = 6

(α + β)² = 6² ⇒ α² + β² + 2αβ = 36 

⇒ α² + β² = 36 – 2 x 6 = 24 

Now, α³ + β³ + α² + β² + α + β 

= (α + β) (α² + β²  - αβ) + (α² + β² ) + (α + β ) 

= 6(24 – 6) + 24 + 6 = 6 × 18 + 30 = 138.

Given α, β, γ are the roots of the equation x³ + ax² + bx + c = 0. Then,

S₁ = α + β + γ = \(-\frac{coefficient\,of\, x^2}{coefficient\,of\,x^3}\) = - a; S₂ = αβ + βγ + αγ = \(-\frac{coefficient\,of\, x^2}{coefficient\,of\,x^3}\) = b

S₃ = αβγ = \(\frac{-constant\,term}{coefficient\,of\,x^3}\) = - c

∴ α^–1 + β^–1 + γ^–1 = \(\frac{1}{α}\) +  \(\frac{1}{β}\) + \(\frac{1}{γ}\) = \(\frac{βγ+αγ+αβ}{αβγ}\) = \(\frac{S_2}{S_3}\) = \(\frac{b}{-c}\) = \(-\frac{b}{c}.\)

Put x = √2

(√2)² + √2(√2) – 4 = 0

2 + 2 – 4

4 – 4

= 0

Therefore, it is the solution to the equation.

When x = –2√2

(–2√2)² + √2(–2√2) – 4 = 0

8 – 4 – 4

= 0

Therefore, it is the solution to the equation.

(b) x² + x – 5 = 0

Same as the previous one. Let’s check the discriminant value for distinct real roots.

a. Given, 2x² - 3√2x + 9/4 = 0

∴ D = b² - 4ac = 0

= (-3√2)² - 4(2) (9/4)

= 18 - 18 = 0

Hence, the roots are real and equal.

b. Given, x² + x – 5 = 0

∴ D = b² - 4ac = 0

= (1)² - 4(1) (-5)

= 1 + 20 = 21 > 0

Hence, the roots are real and distinct.

c. Given, x² + 3x + 2√2 =0

∴ D = b² - 4ac = 0

= 3² - 4(1) (2√2)

= 9 - 8√2 < 0

Hence, the roots are imaginary.

d. Given, 5x² - 3x + 1 = 0

∴ D = b² - 4ac = 0

= (-3)² - 4(5)(1)

= 9 – 20 < 0

Hence, the roots are imaginary.

(a)  [0, 49)

\(5^{56}\big(\frac{1}{5}\big)^x\big(\frac{1}{5}\big)^{\sqrt{x}}\) > 1

⇒ \(5^{56}\times5^{-x}\times5^{-\sqrt{x}}\) < 1 ⇒ \(5^{56-x-\sqrt{x}}\) > 5⁰ 

⇒ 56 - x - √x > 0 ⇒ x + √x - 56 < 0

⇒ y² + y – 56 < 0, where y = √x 

⇒ (y + 8) (y – 7) < 0 ⇒ –8 < y < 7 ⇒ –8 < √x < 7 

⇒ 0 ≤ √x < 7 as √x cannot be negative 

⇒ 0 ≤ \(x\) < 49 ⇒ x ∈ [0, 49)

(a) -7

Given, α, β, γ are the roots of x³ – 2x² + 3x – 4 = 0. Then, 

S₁ = α + β + γ = 2 

S₂ = αβ + βγ + γα = 3 

S₃ = αβγ = 4 

Now, (αβ + βγ + γα)² = α²β² + β²γ² +γ²α² + 2αβ²γ + 2βγ²α + 2α²βγ 

= α²β² + β²γ² +γ²α² + 2αβγ (α + β + γ) 

⇒ α²β² + β²γ² +γ²α² = (αβ + βγ + γα)² – 2αβγ (α + β + γ) 

= 3² – 2 × 4 × 2 = 9 – 16 = – 7.

(b) - 3 and - 4

Let the roots of the quadratic equation x² + px + q = 0 be α and β. According to the given condition, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time, the value of q is correct.

∴ q = Product of roots = αβ = 2 × 6 = 12. 

According to the second condition, B starts with a wrong value of q and obtains the roots as 2 and –9. But this time, the value of p is correct. 

∴ p = sum of roots = α + β = 2 + (–9) = –7           ...(i)

∴ (α - β)² = (α + β)² – 4ab = (–7)² – 4.12 = 49 – 48 = 1 

⇒ α - β = 1                          ...(ii)

∴ Solving equations (i) and (ii), we get α = – 3 and β = – 4.

(c)  – \(\frac{(a^2+b^2)}{2}\)

\(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\)

⇒ \(\frac{(x+b)+(x+a)}{(x+a)(x+b)}\) = \(\frac{1}{c}\) ⇒ \(\frac{2x+b+a}{x^2+(a+b)x+ab}\) = \(\frac{1}{c}\)

⇒ 2cx + (a + b)c = x² + (a + b)x + ab 

⇒ x² + (a + b – 2c)x + ab – ac – bc = 0 

Let α, β be the roots of this equation. Then, 

α + β = – (a + b – 2c) = 0 (Given) 

⇒ a + b = 2c and αβ = ab – ac – bc = ab – (a + b)c

= ab - (a + b) \(\frac{(a+b)}{2}\)

= \(\frac{2ab-(a^2+b^2+2ab)}{2}\) = - \(\bigg(\frac{a^2+b^2}{2}\bigg).\)

C. no real roots

Let’s simplify the equation,

(x² + 1)² - x² = 0

⇒ x⁴ + 2x² + 1 - x² = 0

⇒ x⁴ + x² + 1 = 0

Let x² = y,

⇒ y² + y + 1 = 0

D = b² - 4ac = 0

= 1² - 4(1)(1)

∴ D = -3 < 0

Hence, the equation (x² + 1)² - x² = 0 has no real roots.