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

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

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

D = b² – 4ac = 0

⇒ (2(k + 1))² – 4(3k +1)(k) = 0

⇒ 4(k +1)² – 4(3k² + k) = 0

⇒ (k + 1)² – k(3k + 1) = 0

⇒ 2k² – k – 1 = 0

Now, solving for k by factorization we have

⇒ 2k² – 2k + k – 1 = 0

⇒ 2k(k – 1) + 1(k – 1) = 0

⇒ (k – 1)(2k + 1) = 0,

k = 1 and k = \(\frac{-1}{2}\),

So, the value of k can either be 1 or\(\frac{-1}{2}\).

Given equation,

(x -1) (2x -1) = 0

On expanding it, we get

2x² – 3x + 1 = 0

It is in the form of ax² + bx + c = 0

Where, a = 2, b = -3, c = 1

So, the discriminant is given by D = b² – 4ac

D = (-3)² – 4 x 2 x 1

D = 9 – 8 

= 1

Hence, the discriminant of the given quadratic equation is 1.

Given equation,

x² – 2x + k = 0

It is in the form of ax² + bx + c = 0

Where, a = 1, b = -2, and c = k

So, the discriminant is given by D = b² – 4ac

D = (-2)² – 4(1)(k)

= 4 – 4k

Hence, the discriminant of the given equation is (4 – 4k).

Given equation,

2x² – 5x + 3 = 0

It is in the form of ax² + bx + c = 0

Where, a = 2, b = -5 and c = 3

So, the discriminant is given by D = b² – 4ac

D = (-5)² – 4 x 2 x 3

D = 25 – 24 

= 1

Hence, the discriminant of the given quadratic equation is 1.

Given equation,

x² + 2x + 4 = 0

It is in the form of ax² + bx + c = 0

Where, a = 1, b = 2 and c = 4

So, the discriminant is given by D = b² – 4ac

D = (2)² – 4 x 1 x 4

D = 4 – 16 

= – 12

Hence, the discriminant of the given quadratic equation is – 12.

For a quadratic equation, ax² + bx + c = 0, 

D = b² – 4ac 

If D > 0, roots are real.

(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0

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

⇒ 3x² - 2(a + b + c)x + ab + bc + ac = 0 

⇒ D = 4(a + b + c)² – 12(ab + bc + ac) 

⇒ D = a² + b² + c² + 2ab + 2ac + 2bc – 3ab – 3bc – 3ac 

⇒ D = 1/2 × (2a² + 2b² + 2c² - 2ab – 2ac – 2bc) 

⇒ D = 1/2 × ((a – b)² + (b – c)² + (c – a)²) 

Thus, D is always greater than 0, and the roots are real 

Now, 

when a = b = c, D = 0, 

thus the roots are equal when a = b = c.

Let the breadth of the hall be ‘a’ 

Length = a + 5 

Given, area of the floor of the hall is 84 m² 

⇒ a(a + 5) = 84 

⇒ a² + 5a – 84 = 0 

⇒ a² + 12a – 7a – 84 = 0 

⇒ a(a + 12) – 7(a + 12) = 0 

⇒ (a – 7)(a + 12) = 0 

⇒ a = 7 m 

Length of the hall = 7 + 5 = 12 m

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real.

\(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\)

⇒ D = 4(a + b)² – 8(a² + b²) 

⇒ D = 4a² + 4b² + 8ab – 8a² – 8b² 

⇒ D = - 4(a² + b² – 2ab) = - 4(a – b)² 

Thus, D < 0 for all values of a and b. 

∴ Roots are not real.

For a quadratic equation, ax² + bx + c = 0, 

D = b² – 4ac 

If D = 0, roots are equal 

Given, 

roots of (c² - ab)x² - 2(a² - bc) x + b² - ac = 0 are equal. 

∴ D = 0 

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

⇒ 4(a² – bc)² – 4(c² – ab)(b² – ac) = 0 

⇒ a⁴ + b²c² – 2a²bc – b²c² – a²bc + ab³ + ac³ = 0 

⇒ a(a³ + b³ + c³ – 3abc) = 0 

⇒ a = 0 or a³ + b³ + c³ = 3abc

the given equation is (a - b)x² + 5(a + b)x - 2(a - b) = 0

∴ D = [5(a + b)]² - 4 x (a - b) x [-2(a - b)]

= 25(a + b)² + 8(a - b)²

Since a and b are real and a ≠ b, so (a - b)² > 0 and (a + b)² > 0

∴8(a - b)² > 0 ......... (1) (Product of two positive numbers is always positive)

Also, 25(a + b)² > 0 .......(2) (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25(a + b)² + 8(a - b)² > 0 (Sum of two positive numbers is always positive)

⇒ D > 0

Hence, the roots of the given equation are real and unequal.

Note: For a quadratic equation, ax² + bx + c = 0, we have 

D = b² – 4ac. 

If D = 0, then the roots of the quadratic equation are equal. 

Therefore, 

(p+1)x² - 6(p+1)x + 3(p + 9) = 0 

will have equal roots when, 

⇒ D = 0 

⇒ b² – 4ac = 0 

⇒ b² = 4ac 

Here, 

b = -6(p+1), 

a = (p+1) 

and, c = 3(p+9)

⇒{-6(p + 1)}² = 4×(p + 1)×3(p + 9) 

⇒ 36(p+1)(p+1) = 12(p + 1)(p + 9) 

⇒ 3(p+1)=(p + 9) 

⇒ 3p + 3 - p - 9 = 0 

⇒ 2p - 6 = 0 

⇒ p = 6/2 

⇒ p = 3 

Thus, the value of p is 3 

Now, 

putting the value of p in (p+1)x² - 6(p+1)x + 3(p + 9) = 0, 

we get, 

⇒ 4x² - 24x + 36 = 0 

On taking 4 common, we get, 

⇒ x² – 6x + 9 = 0 

⇒ (x - 3)² = 0 

⇒ x = 3 

Thus, the root of the given equation is x = 3

Let the two consecutive be considered as (x) and (x +1) respectively.

Given that,

The sum of their squares is 85.

Expressing the same by equation we have,

x² + (x + 1)² = 85

⇒ x² + x² + 2x + 1 = 85

⇒ 2x² + 2x + 1 – 85 = 0

⇒ 2x² + 2x – 84 = 0

⇒ 2(x² + x – 42) = 0

Solving for x by factorization method, we get

x² + 7x – 6x – 42 = 0

⇒ x(x + 7) – 6(x + 7) = 0

⇒ (x – 6)(x + 7) = 0

Now, either, x – 6 = 0  ⇒ x = 6

Or, x + 7 = 0 ⇒ x = -7

Thus, the consecutive numbers whose sum of squares can be (6, 7) or (-7, -6).

Let the consecutive numbers be ‘a’ and a + 1. 

Given, sum of squares is 85 

⇒ a² + (a+ 1)² = 85 

⇒ a² + a² + 2a + 1 = 85 

⇒ a² + a – 42 = 0 

⇒ a² + 7a – 6a – 42 = 0 

⇒ a(a + 7) – 6(a + 7) = 0 

⇒ (a – 6)(a + 7) = 0 

⇒ a = 6, -7 

Numbers are, 6, 7 or -7, -6

Let the breadth of the hall be x metres. 

Then the length of the ball is (x + 5) metres.

The area of the floor = x(x + 5) m²

Therefore, x(x + 5) = 84 or x² + 5x - 84 = 0

or (x + 12) (x - 7) = 0 This given x = 7 or x = - 12.

Since, the breadth of the hall cannot be negative, we reject x = - 12 and take x = - only.

Thus, breadth of the hall = 7 metres, and length of the hall = (7 + 5), i.e., 12 metres.

Let’s assume that one part is (x), so the other part will be (29 – x).

From the question, the sum of the squares of these two parts is 425.

Expressing the same by equation we have,

x² + (29 – x)² = 425

⇒ x² + x² + 841 + -58x = 425

⇒ 2x² – 58x + 841 – 425 = 0

⇒ 2x² – 58x + 416 = 0

⇒ x² – 29x + 208 = 0

Solving for x by factorization method, we get

x² – 13x – 16x + 208 = 0

⇒ x(x – 13) – 16(x – 13) = 0

⇒ (x – 13)(x – 16) = 0

Now, either x – 13 = 0 ⇒ x = 13

Or, x – 16 = 0 ⇒ x = 16

Thus, the two parts whose sum of the squares is 425 are 13 and 16 respectively.

Let one of the number be ‘a’. 

Given, sum of two numbers is 29 and the sum of their squares is 425 

⇒ a² + (29 – a)² = 425 

⇒ a² + 841 + a² – 58a = 425 

⇒ a² – 29a + 416 = 0 

⇒ a² – 16a – 13a + 208= 0 

⇒ a(a – 16) – 13(a – 16) = 0 

⇒ (a – 13)(a – 16) = 0 

⇒ a = 13, 16

Let the numbers be ‘a’ and ‘b’. 

Given, sum of two numbers is 48 and their product is 432. 

⇒ a + b = 48 ⇒ a = 48 – b 

Also, 

ab = 432 

⇒ 48b – b² = 432 

⇒ b² – 48b + 432 = 0 

⇒ b² – 36b – 12b + 432 = 0 

⇒ b(b – 36) – 12(b – 36) = 0 

⇒ (b – 12)(b – 36) = 0 

⇒ b = 12, 36

Let the number of boys in the class = x 

Then number of girls in the class = 60 – x [∵ total students = 60] 

Money contributed by the boys = x(60 – x) = 60x – x² [∵ given] 

Money contributed by the girls = (60 – x)x = 60x – x² 

∴ Money contributed by the class = 120x – 2x²

By problem 120x -2x² = 1600 

⇒ 2x² – 120x + 1600 = 0 

⇒ x² – 60x + 800 = 0 

⇒ x² – 40x – 20x + 800 = 0 

⇒ x(x – 40) – 20 (x – 40) = 0 

⇒ (x – 40) (x – 20) = 0 

⇒ x = 40 (or) 20 

∴ Boys = 40 or 20 Girls = 20 or 40.

Correct option is (A) 1 < x < 2

We have \(x=\sqrt{1+\sqrt{1+\sqrt{1\,+......}}}\)

\(\Rightarrow\) \(x=\sqrt{1+x}\)

\(\Rightarrow x^2=1+x\)     (By squaring both sides)

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

\(\Rightarrow x=\frac{-(-1)\pm\sqrt{(-1)^2-4\times1\times-1}}2\)

\(=\frac{1\pm\sqrt5}2\)

\(=\frac{1+\sqrt5}2\) or \(\frac{1-\sqrt5}2\)

\(\because\) \(x=\sqrt{1+\sqrt{1+\sqrt{1\,+......}}}\)

\(\Rightarrow\) x > 1

but \(\frac{1-\sqrt5}2<0<1\)

\(\therefore\) \(x\neq\frac{1-\sqrt5}2\)

Therefore, \(x=\frac{1+\sqrt5}2=\frac{1+2.236}2\)

\(=\frac{3.236}2\)

= 1.618 which is lie between 1 & 2.

\(\therefore\) \(1<x<2\)

Correct option is A) 1 < x < 2

i. Number of trees in a column is x. 

ii. Number of trees in a row = x + 5 

iii. Total number of trees = x x (x + 5) 

iv. According to the given condition,

x(x + 5) = 150 

∴ x² + 5x = 150 

∴ x² + 5x – 150 = 0

v. x² + 15x – 10x – 150 = 0

∴ x(x+ 15) – 10(x + 15) = 0

∴ (x + 15)(x – 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 15 = 0 or x – 10 = 0 

∴ x = -15 or x = 10

But, number of trees cannot be negative.

∴ x = 10

vi. Number of trees in a column is 10. 

vii. Number of trees in a row = x + 5 = 10 + 5 = 15 

∴ Number of trees in a row is 15.

 Correct option is (B) \(35x^2-12x+1=0\)

The equation which is satisfied by \(x=\frac15\) has a root \(x=\frac15.\)

(A) \(2x^2-7x+6=0\)

\(\Rightarrow\) \(2(\frac15)^2-\frac{7}5+6=0\)

\(\Rightarrow\frac{2-35+150}{25}=0\)

\(\Rightarrow\frac{117}{25}=0\)          (Not satisfy)

Hence, \(x=\frac15\) is not a root of equation \(2x^2-7x+6=0.\)

(B) \(35x^2-12x+1=0\)

\(\Rightarrow\) \(35(\frac15)^2-\frac{12}5+1=0\)

\(\Rightarrow\frac{7}{5}-\frac{12}{5}+1=0\)

\(\Rightarrow\frac{-5}{5}+1=0\)

\(\Rightarrow\) -1+1 = 0

\(\Rightarrow\) 0 = 0

Hence, \(x=\frac15\) is a root of \(35x^2-12x+1=0.\)

(C) \(35x^2+12x-1=0\)

\(\Rightarrow\) \(35(\frac15)^2+\frac{12}5-1=0\)

\(\Rightarrow\frac{7}{5}+\frac{12}{5}-1=0\)

\(\Rightarrow\frac{19-5}{5}=0\)

\(\Rightarrow\frac{14}{5}=0\)         (Not satisfies)

Hence, \(x=\frac15\) is not a root of \(35x^2+12x-1=0.\)

(D) \(10x^2-3x-1=0\)

\(\Rightarrow\) \(10(\frac15)^2-\frac{3}5-1=0\)

\(\Rightarrow\frac{2}{5}-\frac{3}{5}-1=0\)

\(\Rightarrow-\frac{1}{5}-1=0\)

\(\Rightarrow\frac{-6}{5}=0\)         (Not satisfies)

Hence, \(x=\frac15\) is not a root of equation \(10x^2-3x-1=0.\) 

Correct option is B) 35x² – 12x + 1 = 0

Correct option is (A) 3

Given that x = 1 is a root of equation \(ax^2+ax+3=0\)

\(\therefore a.1^2+a.1+3=0\)

\(\Rightarrow\) a+a+3 = 0

\(\Rightarrow\) 2a = -3

\(\Rightarrow a=\frac{-3}2\)

Also given that x = 1 is a root of equation \(x^2+x+b=0.\)

\(\therefore1^2+1+b=0\)

\(\Rightarrow b=-2\)

Now, ab \(=\frac{-3}2\times-2=3\)

Correct option is A) 3

(b) real and distinct

The equation is (x – a) (x – a –1) + (x – a – 1) (x – a – 2) + (x – a) (x – a – 2) = 0. 

Let (x – a) = y, then the equation becomes 

y (y – 1) + (y – 1) (y – 2) + y (y – 2) = 0 

⇒ y² – y + y² – 3y + 2 + y² – 2y = 0 ⇒ 3y² – 6y + 2 = 0

∴ Discriminant = D = b² – 4ac = 36 – 4 × 3 × 2 

= 36 – 24 = 12 > 0

∴ Roots are real and distinct.

(d) x² – 9x + 9 = 0.

Equation : x² – (Sum of roots)x + Product of roots = 0 

⇒ x² – 9x + 9 = 0.

(c) \(\frac{3}{4}\)

Let the other root of the equation ax² + x – 3 = 0 be α. 

As (–1) is a root of the given equation, it satisfies the given equation, i.e., a (–1)² + (–1) –3 = 0 ⇒ a – 1 – 3 = 0 ⇒ a = 4. 

∴ The equation becomes 4x² + x – 3 = 0. 

Now, product of roots = α x (–1) = – \(\frac{3}{4}\) 

∴ α = 3 .

(d) a = c

The equation can be written as \(\frac{1}{a}x^2+\frac{1}{b}x+\frac{1}{c}=0\)

i.e., bcx² + acx + ab = 0. 

Let a, \(\frac{1}{α}\) be the roots of the given equation, then 

product of roots = α x \(\frac{1}{α}\) = \(\frac{ab}{bc}\) ⇒ \(\frac{ab}{bc}\) = 1 ⇒ a = c.

Given quadratic equation:

(k+4)x²+(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

∴ D = 0

⇒ (k+1)²−4 × (k+4) × 1=0

⇒ k²+2k+1−4k−16=0

⇒ k²−2k−15=0

⇒k²−5k+3k−15=0

⇒(k−5)(k+3)=0

⇒k−5=0 or k+3=0

⇒k=5 or −3

Thus, the values of k are 5 and −3.

For k = 5:

(k+4)x²+(k+1) x+1=0

⇒ 9x²+6x+1=0

⇒ (3x)²+2(3x)+1=0

⇒ (3x+1)²=0

⇒ x=−1/3, −1/3

For k = −3:

(k+4)x²+(k+1)x+1=0

⇒ x²−2x+1=0

⇒ (x−1)²=0

⇒ x=1,1

Thus, the equal root of the given quadratic equation is either 1 or −13.

A polynomial equation is a quadratic equation, if it is of the form ax² + bx + c = 0 such that a ≠ 0

(i) \(x^{2}+6x-4=0\)

It is a quadratic equation.

(ii) \(\sqrt{3}x^{2}-2x+\frac{1}{2}\)

It is a quadratic equation.

(iii) \(x^{2}+\frac{1}{x^{2}}=5\) 

⇒ x⁴ -5x² + 1 = 0

It is not a quadratic equation as the highest power of x is ‘4’

(iv) \(x-\frac{3}{x}=x^{2}\) 

⇒ x² – 3 = x³

It is not a quadratic equation.

(v) \(2x^{2}-\sqrt{3x}+9=0\)

It is not a quadratic equation as √x is present instead of ‘x’.

(vi) \(x^{2}-2x-\sqrt{x}-5=0\) 

It is not a quadratic equation as an additional √x term is present.

(vii) \(3x^{2}-5x+9=x^{2}-7x+3\) 

⇒ 2x² + 2x + 6 = 0

It is a quadratic equation.

(viii) \(x+\frac{1}{x}=1\)

⇒ x² + 1 – x = 0

It is a quadratic equation.

The quadratic polynomial x² + 6x + 5 can be facorised as follows :-

x² + 6x + 5 = x² + 5x + x + 5

= x(x + 5) + 1 (x + 5) = (x + 5) (x + 1)

Therefore the given quadratic equation becomes (x + 5) (x + 1) =

This gives x = - 5 or = - 1

Therefore, x = - 1 are the required solutions of the given equation.

Obviously, the given equation is valid if x - 3≠0 and 2x + 3≠0.

Multiplying throughout by (x - 3) (2x - 3), we get

2x(2x + 3) + 1(x - 3) + 3x + 9 = 0

or 4x² + 10 + 6 = 0 

or 2x² + 5x + 3 = 0 

or (2x + 3) (x + 1) = 0

But 2x + 3≠ 0, so we get x + 1 = 0. 

This gives x = - 1 as the only solution of the given equation.

Answer is.....

a=3,b=-6

3x² - 2x - 1 = 0

3x² -3x + x - 1 = 0

3x(x -1 ) + 1(x - 1) = 0

(x - 1) (3x + 1) = 0

Either, x - 1 = 0, then x = 1

or 3x + 1 = 0, then x = -1/3

x = 1 or -1/3.

Repeated roots mean d = 0.

d = b² – 4ac

d = (4)² – 4 (9) (6)

d = 16 – 216

d = –200

∴ roots are not repeated.

Repeated roots mean d = 0.

d = b² – 4ac

d = (–6)² – 4(1)(6)

d = 36 – 24

d = 12

∴ roots are not repeated.

Repeated roots mean d = 0.

d = b² – 4ac

d = (–40)² – 4 (16) (25)

d = 1600 – 1600

d = 0

∴ roots are repeated.

Repeated roots mean d = 0.

d = b² – 4ac

d = (–12)² – 4(9)(4)

d = 144 – 144

d = 0

Yes, roots are repeated.

Given as 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]

As we know, i² = –1 ⇒ 1 = –i²

On 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

[On using the formula, a² – b² = (a + b) (a – b)]

(x + 1 + i) (x + 1 – i) = 0

x + 1 + i = 0 or x + 1 – i = 0

x = –1 – i or x = –1 + i

x = -1 ± i

∴ The roots of the given equation are -1 ± i

Given as x² + x + 1 = 0

x² + x + 1/4 + 3/4 = 0

x² + 2 (x) (1/2) + (1/2)² + 3/4 = 0

(x + 1/2)² + 3/4 = 0 [Since, (a + b)² = a² + 2ab + b²]

(x + 1/2)² + 3/4 × 1 = 0

As we know, i² = –1 ⇒ 1 = –i²

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

(x + 1/2)² + 3/4 (-1)² = 0

(x + 1/2)² + 3/4 i² = 0

(x + 1/2)² – (√3i/2)² = 0

[On using the formula, a² – b² = (a + b) (a – b)]

(x + 1/2 + √3i/2) (x + 1/2 – √3i/2) = 0

(x + 1/2 + √3i/2) = 0 or (x + 1/2 – √3i/2) = 0

x = -1/2 – √3i/2 or x = -1/2 + √3i/2

∴ The roots of the given equation are -1/2 + √3i/2, -1/2 – √3i/2

Given as x² – 4x + 7 = 0

x² – 4x + 4 + 3 = 0

x² – 2(x) (2) + 2² + 3 = 0

(x – 2)² + 3 = 0 [Since, (a – b)² = a² – 2ab + b²]

(x – 2)² + 3 × 1 = 0

As we know, i² = –1 ⇒ 1 = –i²

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

(x – 2)² + 3(–i²) = 0

(x – 2)² – 3i² = 0

(x – 2)² – (√3i)² = 0

[On using the formula, a² – b² = (a + b) (a – b)]

(x – 2 + √3i) (x – 2 – √3i) = 0

(x – 2 + √3i) = 0 or (x – 2 – √3i) = 0

x = 2 – √3i or x = 2 + √3i

x = 2 ± √3i

∴ The roots of the given equation are 2 ± √3i

Given as 4x² + 1 = 0

As we know, i² = –1 ⇒ 1 = –i²

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

4x² – i² = 0

(2x)² – i² = 0

[On using the formula, a² – b² = (a + b) (a – b)]

(2x + i) (2x – i) = 0

2x + i = 0 or 2x – i = 0

2x = –i or 2x = i

x = -i/2 or x = i/2

∴ The roots of the given equation are i/2, -i/2

Given as 4x² – 12x + 25 = 0

4x² – 12x + 9 + 16 = 0

(2x)² – 2(2x)(3) + 3² + 16 = 0

(2x – 3)² + 16 = 0 [Since, (a + b)² = a² + 2ab + b²]

(2x – 3)² + 16 × 1 = 0

As we know, i² = –1 ⇒ 1 = –i²

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

(2x – 3)² + 16(–i²) = 0

(2x – 3)² – 16i² = 0

(2x – 3)² – (4i)² = 0

[On using the formula, a² – b² = (a + b) (a – b)]

(2x – 3 + 4i) (2x – 3 – 4i) = 0

2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

2x = 3 – 4i or 2x = 3 + 4i

x = 3/2 – 2i or x = 3/2 + 2i

∴ The roots of the given equation are 3/2 + 2i, 3/2 – 2i

Time = distance/speed 

Given, plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km / hr from its usual speed. 

Let the usual speed be ‘a’.

\(\frac{1600}a{}-\frac{1600}{a\,+\,400}=\frac{40}{60}\)

⇒ 3(1600 × 400) = 2(a² + 400a)

⇒ a² + 400a – 960000 = 0

⇒ a² + 1200a – 800a – 960000 = 0 

⇒ a(a + 1200) – 800(a + 1200) = 0 

⇒ (a + 1200)(a – 800) = 0 

⇒ a = 800 km/hr

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(x-\frac{1}{x-3}=3,x\neq0\)

⇒ x² – 3x – 1 = 3x – 9 

⇒ x² – 6x + 8 = 0 

⇒ x² – 4x – 2x + 8 = 0 

⇒ x(x – 4) – 2(x – 4) = 0 

⇒ (x – 2)(x – 4) = 0 

⇒ x = 2, 4

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(a^2b^2x^2+b^2x-a^2x-1=0\)

⇒ b²x(a²x + 1) – (a²x + 1) = 0 

⇒ (b²x – 1)(a²x + 1) = 0 

⇒ x = -1/a², 1/b²

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\neq2,4\)

⇒ 3(x– 1)(x – 4) + 3(x – 2)(x – 3) = 10(x – 2)(x – 4) 

⇒ 3x² + 12 – 15x + 3x² + 18 – 15x = 10x² – 60x + 80 

⇒ 4x² – 30x + 50 = 0 

⇒ 2x² – 15x + 25 = 0 

⇒ 2x² – 10x – 5x + 25 = 0 

⇒ 2x(x – 5) – 5(x – 5) = 0 

⇒ (2x – 5)(x – 5) = 0 

Thus, x = 5/2, 5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7},x\neq1,-5\)

⇒ 7(x + 5 – x + 1) = 6(x – 1)(x + 5) 

⇒ 42 = 6x² + 24x – 30 

⇒ x² + 4x – 12 = 0 

⇒ x² + 6x – 2x – 12 = 0 

⇒ x(x + 6) – 2(x + 6) = 0 

⇒ (x – 2)(x + 6) = 0 

⇒ x = 2, -6

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(3x^2-2\sqrt{6}x+2=0\)

⇒ (√3x)² – 2√6x + (√2)² = 0 

⇒ (√3x - √2)² = 0

⇒ x \(\frac{\sqrt{2}}{\sqrt{3}},\)\(\frac{\sqrt{2}}{\sqrt{3}}\)

For roots to be real its D ≥ 0

\(\sqrt{(m+1)^2 - 4(1)(m+4)}≥ 0\) 

(m + 1)² – 4(m + 4) ≥ 0 

m² – 2m – 15 ≥ 0 

(m – 1)² – 16 ≥ 0 

(m – 1)² ≥ 16

m – 1 ≤ -4 or m – 1 ≥ 4 

m ≤ -3 or m ≥ 5 

For both roots to be negative product of roots should be positive and sum of roots should be negative. 

Product of roots = m + 4 > 0 

⇒ m > -4 

Sum of roots = m + 1 < 0 

⇒ m < -1 

After taking intersection of D ≥ 0, Product of roots > 0 and sum of roots < 0. We can say that the final answer is 

m ∈ (-4, -3]

Given, \(\frac{(x+2)(x-5)}{(x-3)(x+6)} = \frac{x-2}{x+4} \) 

(x+2) (x-5) (x+4) = (x-2) (x-3) (x+6) 

x³ + 4x² - 5x² - 20x + 2x² + 8x - 10x - 40 = x³+6x²-3x²-18x- 2x² - 12x + 6x + 36 

x²- 22x - 40 = x² - 24x + 36 

4x = 76 

x = 19 

hence the given equation has only one solution.

Given 4x²+ 3x + 7 = 0 

We know sum of the roots = \(\frac{-3}{4}\)

Product of the roots =  \(\frac{7}{4}\)

As given that α and β are roots then,

α + β  =  \(\frac{-3}{4}\) 

α β =  \(\frac{7}{4} \)  

Given \(\frac{1}{α} + \frac{1}{β}\) 

= \(\frac{α+β}{αβ}\) 

= \(-\frac{3}{\frac{4}{\frac{7}{4}}}\) 

= \(\frac{-3}{7}\)