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Since the roots of the given equation are real and distinct, we must have D > 0

=> 64 - 4 (a² - 6a) > 0 

=> 4[16 - a² + 6a] > 0 

=> -4(a² - 6a - 16) > 0

=> a² - 6a - 16 < 0 

=> (a - 8) (a + 2) < 0 

=> - 2 < a < 8

Hence, the roots of the given equation are real if ‘a’ lies between -2 and 8.

Let x be one of the positive integers. Then the other integer is x + 1, x∈Z⁺

Since the sum of the squares of the integers is 545, we get

x² + (x + 1)² = 545

or 2x² + 2x - 544 = 0

or x² + x - 272 = 0

x² + 17x - 16x - 272 = 0

or x(x + 17) - 16(x + 17) = 0

or (x - 16) (x + 17) = 0

Here, x = 16 or x = -17 But, x is a positive integer. Therefore, reject x = - 17 and take x = 16. Hence, two consecutive positive integers are 16 and (16 + 1), i.e., 16 and 17.

Correct option is (D) 45

We have \(\sqrt{x^2-5x-1}=2\)

\(\Rightarrow x^2-5x-1=2^2=4\)        (By squaring both sides)

\(\Rightarrow x^2-5x-1-4=0\)

\(\Rightarrow x^2-5x-5=0\)

Discriminant \(=b^2-4ac\)

\(=(-5)^2-4\times1\times-5\)

= 25+20 = 45

Correct option is D) 45

From the given information, we have:

n (n + 2) = 168

n² + 2n − 168 = 0

n² + 14n − 12n − 168 = 0

n(n + 14) − 12(n + 14) = 0

(n + 14) (n − 12) = 0

n = −14, 12

But, n cannot be negative.

Therefore, n = 12.

(b) 3abc

Given that the roots are real and equal, 

D =0 ⇒ b² – 4ac = 0 for ax² + bx + c = 0.

∴ [–2(a² – bc)]² – 4(c² – ab) (b² – ac) = 0

⇒ 4[a⁴ + b²c² – 2a²bc – c²b² + ac³ + ab³ – a²bc) = 0 

⇒ 4a (a³ + b³ + c³ – 3 abc) = 0 

⇒ a³ + b³ + c³ = 3abc.

(c) real, rational and unequal.

Given equation is x² – 2√3x – 22 = 0. 

Discriminant = D = b² – 4ac (for ax² + bx + c = 0) 

= (–2√3)² – 4(–22) = 12 + 88 = 100

As D > 0 and is a perfect square, the roots are real, rational and unequal.

(b) a² – b² + 2ac = 0.

As sin θ and cos θ are the roots of the equation ax² – bx + c = 0.

∴ sin θ + cos θ = \(\frac{b}{a}\) and sin θ cos θ = \(\frac{c}{a}\)

⇒ (sin θ + cos θ)² = \(\frac{b^2}{a^2}\)

⇒ sin²θ + cos²θ + 2sin θ + cos θ = \(\frac{b^2}{a^2}\)

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

⇒ 2ac = b² – a² ⇒ a² – b² + 2ac = 0.

2x² - 5x + 3 = 0

2x²/2 - 5x/2 + 3/2 = 0 (dividing both sides by 2)  

x² - 5x/2 + 3/2 = 0

[x² - 5x/2 + (5/4)²] - (5/4)² + 3/2 = 0

[x - 5/4]² - 25/16 + 3/2 = 0

[x - 5/4]² - (25-24)/16 = 0

[x - 5/4]² - 1/16 = 0

[x - 5/4]² = 1/16

x - 5/4 = ± 1/4   (square rooting on both sides)  

x = 1/4 + 5/4  or x = -1/4 + 5/4

x = (1+5)/4   or x = (-1+5)/4

x = 6/4   or x = 4/4

x = 3/2  or  x = 1

Given, two numbers differ by 4. 

Let one of the numbers be ‘a’. 

Second number = a – 4 

Also, their product is 192. 

⇒ a(a – 4) = 192 

⇒ a² – 4a – 192 = 0 

⇒ a² – 16a + 12a – 192 = 0 

⇒ a(a – 16) + 12(a – 16) = 0 

⇒ (a + 12)(a – 16) = 0 

⇒ a = -12, 16 

Thus numbers are -12, -16 or 12, 16

x² - 4x + 1 = 0

⇒ x² - 4x = -1

⇒ x² - 2 \(\times\)x \(\times\)2 + 2² = - 1 + 2²  (Adding 2² on both sides)

⇒  (x - 2)² = - 1 + 4 = 3

⇒ x - 2 = ±√3

⇒ x - 2 = √3 or x - 2 = - √3

⇒ x = 2 + √3 or x = 2 -√3

Hence, 2 + √3 and 2 - √3 are the roots of the given equation.

x² - 6x + 3 = 0

⇒ x² - 6x = - 3

⇒ x² - 2 \(\times\)x \(\times\)3 + 3² = - 3 + 3²   (Adding 3² on both sides)

⇒ (x - 3)² = - 3 + 9 = 6

⇒ x  - 3 = ± √6  (Taking square root on the both sides)

⇒ x - 3 = √6 or x - 3 = - √6

⇒ x = 3 + √6 or x = 3 - √6

Hence, 3 + √6 and 3 - √6 are the roots of the given equation.

Correct option is (A) either k ≥ 2 or k ≤ -2

If expression \(x^2+kx+1\) is factorizable into two linear factors then \(k^2-4\times1\times1\geq0\)   \((\because b^2-4ac\geq0\) for two linear factors)

\(\Rightarrow\) \(k^2-4\geq0\)

\(\Rightarrow\) \(k^2\geq4\)

\(\Rightarrow\) either \(k\geq2\) or \(k\leq-2\)

Correct option is A) either k ≥ 2 or k ≤ -2

Correct option is (B) x, x + 1

Two consecutive positive integers are differ by 1.

\(\therefore\) x & (x+1) are two consecutive positive integers.

Correct option is B) x, x + 1

Correct option is (B) 11 and 13

Let x & x+2 are two consecutive odd numbers.

\((\because\) Difference between two consecutive odd numbers is 2)

\(\therefore x\times(x+2)=143\)     \((\because\) Given that product of two consecutive odd integers is 143)

\(\Rightarrow x^2+2x-143=0\)

\(\Rightarrow x^2+13x-11x-143=0\)

\(\Rightarrow x(x+13)-11(x+13)=0\)

\(\Rightarrow(x+13)(x-11)=0\)

\(\Rightarrow\) x+13 = 0 or x - 11 = 0

\(\Rightarrow\) x = -13 or x = 11

\(\therefore\) x+2 = -13+2 = -11

or x+2 = 11+2 = 13

Hence, required odd numbers are 11 and 13.

Correct option is B) 11 and 13

Correct option is (B) 17m and 5m

Let \(l\;and\;b\) are parameters of rectangular room.

\(\therefore2(l+b)=34\)        \((\because\) Perimeter = 34 m (given))

\(\Rightarrow l+b=\frac{34}2=17\)   _____________(1)

\(\because\) Length of diagonal = 13

\(\Rightarrow\sqrt{l^2+b^2}=13\)

\(\Rightarrow l^2+b^2=13^2=169\)   _____________(2)

Now, \((l+b)^2=l^2+b^2+2lb\)

\(\Rightarrow17^2=169+2lb\)      (From (1) & (2))

\(\Rightarrow2lb=289-169\)

\(\Rightarrow lb=\frac{120}2=60\)   _____________(3)

Now, \((l-b)^2=(l+b)^2-4lb\)

\(=17^2-4\times60\)          (From (1) & (3))

= 289 - 240

= 49 \(=7^2\)

\(\Rightarrow l-b=7\)   _____________(4)

By adding (1) & (4), we get

\((l+b)+(l-b)=17+7\)

\(\Rightarrow2l=24\)

\(\Rightarrow l\) = 12 m

Put \(l=12\) in equation (1), we get

12+b = 17

\(\Rightarrow\) b = 17 - 12 = 5 m

Hence, dimensions of the room are 17m and 5m.

Correct option is B) 17m and 5m

Correct option is (C) ± 1

Let \(\alpha\) be common root of equations

\(x^2+4ax+3=0\) and \(2x^2+3ax-9=0\)

\(\therefore\) \(\alpha^2+4a\alpha+3=0\)     _______________(1)

\(2\alpha^2+3a\alpha-9=0\)       _______________(2)

Multiply equation (1) by 2, we get

\(2\alpha^2+8a\alpha+6=0\)     _______________(3)

Subtract equation (2) from (3), we get

\((2\alpha^2+8a\alpha+6)-(2\alpha^2+3a\alpha-9)\) \(=0-0=0\)

\(\Rightarrow5a\alpha+15=0\)

\(\Rightarrow a\alpha=\frac{-15}5=-3\)     _______________(4)

Put \(a\alpha=-3\) into equation (1), we get

\(\alpha^2-12+3=0\)

\(\Rightarrow\) \(\alpha^2-9=0\)

\(\Rightarrow\) \(\alpha^2=9\)

\(\Rightarrow\) \(\alpha=\pm3\)

Then from (4), we get

\(a=\frac{-3}\alpha=\frac{-3}{\pm3}=\mp1\)

\(\therefore\) a = \(\pm\,\,1\)

Correct option is C) ± 1

Here, Factors of constant term (a² - b²) are (a - b) and (a + b).

Also, Coefficient of the middle term = - 2a = - [(a - b) + (a + b)]

So, x² - 2ax + a² - b² = 0

=> x² - {(a - b) + (a + b)}x + (a - b)(a + b) = 0  

=> x² - (a - b) x - (a + b) x + (a - b) (a + b) 

=> x{x - (a - b)}- (a + b) {x - (a - b)} = 0 

=> {x - (a - b)} {x - (a + b)} = 0

=> x - (a - b) = 0 or, x - (a + b) = 0 

=> x = a - b or x = a + b

The general form of quadratic equation is ax² + bx + c = 0, where a,b,c are real numbers and a≠0 Since a≠0 , quadratic equations, in general are of the following types :-

(i) b = 0, c≠0 i.e., of he type ax² + c=0. (ii) b ≠0, c = 0, i.e. of the type ax² + bx = 0.

(iii) b = 0, c = 0, i.e. of the type ax² = 0. (iv) b≠0 , c ≠ 0, i.e., of the type ax² + bx + c = 0.

If P(x) is a quadratic expression in variable x, then P(x) = 0 is known as a quadratic equation.

i. y² – 7y + 12 = 0 

ii. x² – 8 = 0

i. 2y – 10 – y² 

∴ y² + 2y – 10 = 0

Comparing the above equation with

ay² + by + c = 0, we get 

a = 1, b = 2, c = -10

ii. (x – 1)² = 2x + 3 

∴ x² – 2x + 12x + 3 

x² – 2x + 1 – 2x – 30 

∴ x² – 4x – 2 = 0

Comparing the above equation with

ax² + bx + c = 0, we get 

a = 1, b = -4, c = -2

iii. x² + 5x = – (3 – x) 

∴ x² + 5x = -3 + x 

∴ x² + 5x – x + 3 = 0 

∴ x^{​​​​​​​2} + 4x + 3 = 0

Comparing the above equation with

ax² + bx + c = 0, we get 

a = 1, b = 4, c = 3

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.

iii. The given equation is

m³ + 3m² – 2 = 3m 

∴ 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. 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.

i. In the equation 9y² + 5 = 0, [y] is the only variable and maximum index of the variable is [2]. 

∴ It [is] a quadratic equation.

ii. In the equation m³ – 5m² + 4 = 0, [m] is the only variable and maximum index of the variable is not 2.

∴ It [is not] a quadratic equation.

iii. (l + 2)(l – 5) = 0 

∴ l(l – 5) + 2(l – 5) = 0 

∴ l² – 5l + 2l – 10 = 0 

∴ l² – 3l – 10 = 0.

In this equation [l] is the only variable and maximum index of the variable is [2]

∴ it [is] a quadratic equation.

The following table is

Let the given Q.E. be p(x) = 2x² – 5x + 3 

Now p(1) = 2(1)² – 5(1) + 3 

= 2 – 5 + 3 = 0 

∴ 1 is a root of 2x² – 5x + 3 = 0

also p \((\frac{3}{2})\) = 2 \({(\frac{3}{2})}^2\) – 5 \((\frac{3}{2})\) + 3 

= 2 × \(\frac{9}{4}\) – \(\frac{15}{2}\) + 3 

= \(\frac{9}{2}\) + 3 – \(\frac{15}{2}\)

= \(\frac{9+6-15}{2}\)= 0 

∴ \(\frac{3}{2}\) is also a root of 2x² – 5x + 3 = 0.

Index form of the given polynomials:

x² + 3x – 5, 3x² – 5x + 0, 5x² + 0x + 0

i. Coefficients of x² are [1], [3] and [5] respectively, and these coefficients are non zero.

ii. Coefficients of x are 3, [-5] and [0] respectively. 

iii. Constant terms are [-5], [0] and [0] respectively.

Here, constant terms of second and third polynomial is zero.

Given x² + 2x + 2 = 0 

⇒ x²+ 2x + 1 + 1 = 0 

⇒ x² + 2(x)(1) + 1² + 1 = 0 

⇒ (x + 1)² + 1 = 0 [∵ (a + b)² = a² + 2ab + b²] 

We have i² = –1 

⇒ 1 = –i² 

By substituting 1 = –i² in the above equation, we get 

(x + 1)² + (–i²) = 0 

⇒ (x + 1)² – i² = 0 

⇒ (x + 1)² – (i)² = 0 

⇒ (x + 1 + i)(x + 1 – i) = 0 [∵ a² – b² = (a + b)(a – b)] 

⇒ x + 1 + i = 0 or x + 1 – i = 0

⇒ x = –1 – i or x = –1 + i 

∴ x = –1 ± i 

Thus, the roots of the given equation are –1 ± i.

Let breadth of the rectangle = x 

Length = x + 2 ,

Area = 120 sq. units 

x(x+2) = 120 

x² + 2x- 120 = 0 

(x + 12) (x- 10) = 0 

x = – 12 or 10 

Breadth cannot be negative 

∴ Breadth of the rectangle = x 

= 10 units 

∴ Length = x + 2 = 10 + 2 = 12 units

Let,

Pauls's present age be x.

So his father's age will be 2(x²)

A/Q,

2x² +10 = 4(x+10)

2x² + 10 = 4x + 40

2x² - 4x = 40-10

2x² -4x = 30

2x² -4x - 30 = 0

2x² - 10x + 6x - 30 = 0

2x(x-5) + 6(x-5) = 0

(2x + 6) (x-5) = 0

2x + 6 = 0  or  x - 5 = 0

2x = -6     or x = 5

x = -6/2  or x = 5

x = -3  or  x = 5

So x = -3, 5

So the age should be in negative value therefore present age of paul is 5 years and his father's age 2(5)² = 2 x 25 = 50 years.

The given equation 4x² – 3kx + 1 = 0 is in the form of ax² + bx + c = 0

Where a = 4, b = -3k, c = 1

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (-3k)² – 4(4)(1) = 0

⇒ 9k² – 16 = 0

⇒ k = ± \(\frac{4}{3}\)

The value of k is ± \(\frac{4}{3}\).

Given,

4x² + kx + 3 = 0

It’s of the form of ax² + bx + c = 0

Where, a = 4, b = k, c = 3

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

D = (k)² – 4(4)(3) ≥ 0

⇒ k² – 48 ≥ 0

⇒ k² ≥ 48

⇒ k ≥ 4√3 and k ≤ -4√3 [After taking square root on both sides]

The value of k can be represented as (∞, 4√3] U [-4√3, -∞)

The given equation kx² + kx + 1 = -4x² – x

This can be rewritten as,

(k + 4)x² + (k + 1)x + 1 = 0

Now, this in the form of ax² + bx + c = 0

Where a = (k +4), b = (k + 1), c = 1

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (k + 1)² – 4(k +4)(1) = 0

⇒ (k +1)² – 4k – 16 = 0

⇒ k² + 2k + 1 – 4k – 16 = 0

⇒ k² – 2k – 15 = 0

Now, solving for k by factorization we have

⇒ k² – 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

⇒ (k + 3)(k – 5) = 0,

k = -3 and k = 5,

So, the value of k can either be -3 or 5.

The given equation kx² + 4x + 1 = 0 is in the form of ax² + bx + c = 0

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

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ 4² – 4(k)(1) = 0

⇒ 16 – 4k = 0

⇒ k = 4

The value of k is 4.

The given equation 9x² – 24x + k = 0 is in the form of ax² + bx + c = 0

Where a = 9, b = -24, c = k

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (-24)² – 4(9)(k) = 0

⇒ 576 – 36k = 0

⇒ k = 16

The value of k is 16.

The given equation 4x² + kx + 9 = 0 is in the form of ax² + bx + c = 0

Where a = 4, b = k, c = 9

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ k² – 4(4)(9) = 0

⇒ k² – 144 = 0

⇒ k = ± 12

The value of k is 12 or -12.

The given equation 2kx² – 40x + 25 = 0 is in the form of ax² + bx + c = 0

Where a = 2k, b = -40, c = 25

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (-40)² – 4(2k)(25) = 0

⇒ 1600 – 200k = 0

⇒ k = 8

The value of k is 8.

Let the numbers be ‘a’ and ‘b’. 

Given, sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. 

⇒ a + b = 8 

⇒ a = 8 – b 

Also, 

15 × (1/a + 1/b) = 8 

⇒ 1/a + 1/b = 8/15 

⇒ 1/a + 1/(8 – a) = 8/15 

⇒ 15(8 – a + a) = 8(8a – a²) 

⇒ a² – 8a + 15 = 0 

⇒ a² -5a – 3a + 15 = 0 

⇒ a(a – 5) -3(a – 5) = 0 

⇒ (a – 3)(a – 5) = 0 

⇒ a = 3, 5

Let the number of swans in the pond be ‘a’.

Given, out of a group of swans, 7/2 times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water.

\(\Rightarrow \frac{7}{2}\sqrt{a}+2=a\)

⇒ 7√a = 2a – 4

Squaring both sides

⇒ 49a = 4a² + 16 – 16a 

⇒ 4a2 – 65a + 16 = 0 

⇒ 4a² – 64a – a + 16 = 0 

⇒ 4a(a – 16) – (a – 16) = 0 

⇒ (4a – 1)(a – 16) = 0 

⇒ a can’t be 1/4, 

thus a = 16

(a) 16

Let the two little bands of monkeys have x monkeys. Then,

\(\big(\frac{x}{8}\big)^2 + 12 = x\) ⇒ \(\frac{x^2}{64} + 12 = x\)

⇒ x² - 64x + 768 = 0

⇒ (x - 16)(x - 48) = 0

⇒ x = 16, 48

Seeing the options, x = 16.

Let the number be x.

So, its square will be x².

From the question, it’s given that sum of the number and its square is \(\frac{63}{4}\)

Which means,

x + x² = \(\frac{63}{4}\)

⇒ 4x + 4x² = 63

⇒ 4x² + 4x – 63 = 0

Solving for x by factorization method, we have

⇒ 4x² + 18x – 14x – 63 = 0

⇒ 2x(2x + 9) – 7(2x – 9) = 0

⇒ (2x – 7)(2x + 9) = 0

Now, either 2x -7 = 0 ⇒ x = \(\frac{7}{2}\)

Or, 2x + 9 = 0 ⇒ x = \(\frac{-9}{2}\)

Thus, the numbers are \(\frac{7}{2}\) and \(\frac{-9}{2}\).

(d) 5 m

Let the length of one side be x m. 

∴ Area of the plot = 60 m, length of the other side = \(\frac{60}{x}\)

Given, 2\(\big(x+\frac{60}{x}\big) = 34\)

⇒ x² + 60 = 17x

⇒ x² - 17x + 60 = 0

⇒ x² - 12x - 5x + 60 = 0

⇒(x - 12)(x - 5) = 0

⇒ x = 12, 5

∴ Length of each shorter side = 5 m.

(b) Rs 12

Let the present price be Rs k per dozen.

∴ Price per unit = Rs \(\frac{k}{12}\)

∴ Increased price = Rs (k +2) per dozen

∴ Increase price per unit = Rs \(\frac{(k+2)}{12}\)

Given, \(\frac{56}{\frac{k}{12}}-\frac{56}{\frac{(k +2)}{12}} = 8\)

⇒ \(\frac{12\times56}{k}-\frac{12\times56}{k+2}=8\)

⇒ 672 x (k + 2) - 672k = 8k(k + 2)

⇒ 8k² + 16k - 1344 = 0

⇒ k² + 2k - 168 = 0 ⇒ (k + 14)(k - 12) = 0

⇒ k = 12

Let the consecutive even integers be ‘a’ and a + 2 

⇒ a² + (a + 2)² = 340 

⇒ 2a² + 4a – 336 = 0 

⇒ a² + 2a – 168 = 0 

⇒ a² + 14a – 12a – 168 = 0 

⇒ a(a + 14) – 12(a + 14) = 0 

⇒ (a – 12)(a + 14) = 0 

Thus, a = 12 or – 14 

Consecutive even integers are 12, 14 or -14, - 12

Let the numbers be ‘a’ and ‘b’. 

Given, difference of two numbers is 4 and difference of their reciprocals is \(\frac{4}{21}\)

⇒ a – b = 4 

⇒ a = b + 4 and 1/b – 1/a = 4/21 

⇒ 1/(b + 4) – 1/b = -4/21 

⇒ 21(b – b – 4) = -4(b² + 4b) 

⇒ b² + 4b - 21 = 0 

⇒ b² + 7b – 3b – 21 = 0 

⇒ b(b + 7) – 3(b + 7) = 0 

⇒ (b – 3)(b + 7) = 0 

⇒ b = 3, - 7 

Numbers are , 3, 7 or -7, -3

Let the first number be ‘X’, so the other number will be ’(12–X)’.

∵ X (12 – X) = 32

12X – X² = 32

X² – 12X + 32 = 0

On factorising further,

X² – 4X – 8X + 32 = 0

X(X – 4) – 8(X – 4) = 0

(X – 4)(X – 8) = 0

So, X = 4 or 8

∴ The numbers are 4 and 8.

x²(a² + b²) + 2x(ac + bd) + (c² + d²)= 0

d = b² – 4ac

d = (2ac + 2bd)² – 4 (a² + b²) (c² + d²)

d = 4a²c² + 4b²d² + 8abcd – 4 [a² (c² + d²) + b² (c²+d²)]

d = 4a²c² + 4b²d² + 8abcd – 4 [a²c² + a²d² + b²c² + b²d²]

d = 4a²c² + 4b²d² + 8abcd – 4a²c² – 4a²d² – 4b²c² – 4b² d²

d = 8abcd – 4a²d² – 4b²c²

d = 8abcd – 4(a²d² + b² c²)

d = –4 (a² d² + b²c² – 2abcd)

d = –4 [(ad + bc)²]

For ad ≠ bc

d= –4 × [value of (ad + bc)²]

∴ d is always negative

So, d < 0

The given equation has no real roots.

Roots are equal

∴ d = 0

d = b² – 4ac

d = (4k)² – 4 (k² – k + 2) (1)

d = 16k² – 4k² + 4k – 8

Put d = 0

0 = 16k² – 4k² + 4k – 8

12k² + 4k² + 4k – 8 = 0 (divide by 4)

3k² + k – 2 = 0

3k² + 3k – 2k – 2 = 0

3k (k + 1) – 2k (k + 1) = 0

(3k–2k)(k+1) = 0

(3k–2k)=0 or (k+1) = 0

k = 0 k = –1

The given equation (k +1)x² + 2(k +3)x + (k +8) = 0 is in the form of ax² + bx + c = 0

Where a = (k +1), b = 2(k + 3), c = (k + 8)

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (2(k + 3))² – 4(k +1)(k + 8) = 0

⇒ 4(k +3)² – 4(k² + 9k + 8) = 0

⇒ (k + 3)² – (k² + 9k + 8) = 0

⇒ k² +6k + 9 – k² – 9k – 8 = 0

⇒ -3k + 1 = 0

⇒ k = \(\frac{1}{3}\)

So, the value of k is \(\frac{1}{3}\).

Given,

kx(x – 3) + 9 = 0

It can be rewritten as,

kx² – 3kx + 9 = 0

It’s of the form of ax² + bx + c = 0

Where, a = k, b = -3k, c = 9

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

D = (-3k)² – 4(k)(9) ≥ 0

⇒ 9k² – 36k ≥ 0

⇒ 9k(k – 4) ≥ 0

⇒ k ≥ 0 and k ≥ 4

⇒ k ≥ 4

The value of k should be greater than or equal to 4 to have real roots.