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Let Nishu take x days to complete the work alone. 

∴ Total work done by Nishu in 1 day = \(\frac{1}{x}\) Also, Pintu takes (x + 6) days to complete the work alone.

∴ Total work done by Pintu in 1 day =\(\frac{1}{x+6}\)

∴ Total work done by both in 1 day = (\(\frac{1}{x}\) +  \(\frac{1}{x+6}\))

But, both take 4 days to complete the work together. 

∴ Total work done by both in 1 day = 1/4

According to the given condition ,

1/x + 1/(x+6) = 1/4

∴ x + (6+x) / x(x+6) = 1/4

∴ 2x + 6 / x(x+6) = 1/4

∴ 4(2x + 6) = x(x + 6) 

∴ 8x + 24 = x + 6x 

∴ x² + 6x – 8x – 24 = 0 

∴ x² – 2x – 24 = 0 

∴ x² – 6x + 4x – 24 = 0 

∴ x(x – 6)+ 4(x – 6) = 0 

∴ (x – 6) (x + 4) = 0

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

∴ x – 6 = 0 or x + 4 = 0 

∴ x = 6 or x = -4 

But, number of days cannot be negative, 

∴ x = 6 and x + 6 = 6 + 6 = 12

∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

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.

The correct answer is : x (x + 5) = 2

i. 3y² + kg + 12 = 0

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

a = 3, b = k, c = 12 

∴ ∆ = b² – 4ac 

= (k)² – 4 × 3 × 12 

= k² – 144 = k² – (12)² 

∴ ∆ = (k + 12) (k – 12)    …[∵ a – b = (a + b) (a – b)]

Since, the roots are real and equal. 

∴ ∆ = 0 

∴ (k + 12) (k – 12) = 0 

∴ k + 12 = 0 or k – 12 = 0 

∴ k = -12 or k = 12

ii. kx (x – 2) + 6 = 0 

∴ kx² – 2kx + 6 = 0 

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

a = k, b = -2k, c = 6 

∴ ∆ = b – 4ac 

= (-2k)² – 4 × k × 6

= 4k² – 24k 

∴ ∆ = 4k (k – 6) 

Since, the roots are real and equal. 

∴ ∆ = 0 

∴ 4k (k – 6) = 0 

∴ k(k – 6) = 0 

∴ k = 0 or k – 6 = 0 But, if k = 0 then quadratic coefficient becomes zero. 

∴ k ≠ 0

∴ k = 6

The correct answer is : (A) x² + 4x = 11 + x²

The correct answer is : (C) 0 or 4

The correct answer is : (B) x² – 8x + 15 = 0

The correct answer is : (B) 17

(2x + 3)² = 81

⟹ 2x + 3 = ± 9

⟹ 2x + 3 = 9 and 2x + 3 = −9

⟹ 2x = 6 and 2x = − 12

⟹ x = 3 and x = − 6

x (2x + 5) = 3 ⇒ 2x² + 5x = 3

⇒ 2x² + 5x - 3 = 0 ⇒ 2x² + 6x - x - 3 = 0

⇒ 2x (x + 3) - 1 (x + 3) = 0

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

⇒ x = \(\frac12\) or x = - 3

∴  x = \(\frac12\), -3.

\(\frac{x^2-4}{3} = 20\)

⇒ x² - 4 = 60 ⇒ x² - 64 = 0 ⇒ (x + 8)(x - 8) = 0

⇒ (x + 8) = 0 or (x - 8) = 0 ⇒ x = - 8 or 8

∴ x = -8, 8

x² - 3x -10 = 0

⇒ x² -5x + 2x - 10 = 0

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

⇒ x - 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

∴ The roots of the quadratic equation x² - 3x -10 = 0 are 5 and -2.

A polynomial of the form ax² bx + c where a ≠ 0 and a, b, c are real constants and x is a real variable is called a quadratic polynomial.

Here ,  a = 15     ,   b = 1    ,   c = 1/4

We know that , D = b²- 4ac

D = 1² - 4 × 15 × 1/4 

D  =1 - 15       =   - 14 

Since , D < 0   This quadratic equation doesn't have real roots .

Given x² + 1 = 0 

We have i² = –1 ⇒ 1 = –i² 

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

x² – i² = 0 

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

⇒ x + i = 0 or x – i = 0 

⇒ x = –i or x = i 

∴ x = ±i

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

1. a) 2(x + 5)km

2. c) 25km/ hour

3. a) 20km/ hour

4. d) 16 hour

1. c) (20 – x)km/hr

2. b) distance=(speed)/ time

3. c) x² + 30x – 400= 0

4. b) 10 km/hour

5. c) 45 minute

x + \(\frac{6}{x}\) = 7 ⇒ \(\frac{x^2+6}{x}\) = 7 

⇒ x² – 7x + 6 =0 

⇒ (x – 6) (x – 1) = 0 

⇒ x = 6 or 1 

Roots = 6, 1

Here we have,

LHS = x² − 3√3x + 6

Substituting x = √3 in LHS, we get

(√3)² − 3√3(√3) + 6

⇒ 3 – 9 + 6 = 0 = RHS

⇒ LHS = RHS

Thus, x = √3 is a solution of the given equation.

Similarly,

Substituting x = −2√3 in LHS, we get

(-2√3)² − 3√3(-2√3) + 6

⇒ 12 + 18 + 6 = 36 ≠ RHS

⇒ LHS ≠ RHS

Thus, x = −2√3 is not a solution of the given equation.

Put both the values of x in the equation.

When x = √3

(√3)² –3√3(√3) + 6 = 0

3 – (3)3 + 6

3 – 9 + 6

9 – 9

= 0

Therefore, it is the solution to the equation.

When x = –2√3

(–2√3)² –3√3(–2√3) + 6 = 0

12 + 18 + 6

= 36

Therefore, it is not the solution to the equation.

x² + 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b² – 4ac = 16 – 20 = – 4 < 0.

Since the square of a real number cannot be negative, therefore √b²-4ac not have any real value.

So, there are no real roots for the given equation.

Let the numbers be x and x – 3, as its given the number differ by 3.

From the question, it’s given that sum of squares of these numbers is 117.

x² + (x – 3)² = 117

⇒ x² + x² + 9 – 6x – 117 = 0

⇒ 2x² – 6x – 108 = 0

⇒ x² – 3x – 54 = 0

Solving for x by factorization method, we have

⇒ x² – 9x + 6x – 54 = 0

⇒ x(x – 9) + 6(x – 9) = 0

⇒ (x – 9)(x + 6) = 0

Now, either x – 9 = 0 ⇒ x = 9

Or, x + 6 = 0 ⇒ x = – 6

Considering only the positive value of x as natural numbers are always positive i.e, x = 9.

So, x – 3 = 6.

Thus, the two numbers are 6 and 9 respectively.

Here ,

The given equation is ;

sinx + cosx = x² - 2x + √35 .

Here ,

=> LHS = sinx + cosx

=> LHS = √2[ sinx•(1/√2) + cosx•(1/√2) ]

=> LHS = √2[ sinx•sin45° + cosx•sin45° ]

=> LHS = √2sin(x + 45°)

Also ,

We know that , -1 ≤ sin∅ ≤ 1 .

Thus ,

=> -1 ≤ sin(x + 45°) ≤ 1

=> -1•√2 ≤ √2•sin(x + 45°) ≤ 1•√2

=> -√2 ≤ √2sin(x + 45°) ≤ 2

=> -√2 ≤ LHS ≤ √2

Now ,

=> RHS = x² - 2x + √35

=> RHS = x² - 2x + 1² - 1² + √35

=> RHS = (x - 1)² + 1 + √35

Also ,

We know that , x² ≥ 0 .

Thus ,

=> (x - 1)² ≥ 0

=> (x - 1)² + 1 + √35 ≥ 1 + √35

=> RHS ≥ 1 + √35

Observing LHS and RHS , we can conclude that there exist no real number for which LHS and RHS would be equal . Thus , there is no real solution of the given equation .

M³ - 3M² - 6M = 0

M(M² - 3M - 6) = 0

M = 0  or M² - 3M - 6 = 0 ....(i)

Solving (i) by completing the square method,

M² - 3M - 6 = 0

M = {3 ±√9-4(1)(-6)} / 2

M = (3 ±√9+24) / 2

M = (3 ±√33) / 2

So, the roots are M = 0 or M = M = (3 ±√33) / 2

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.

\(\frac{x+3}{x+2}=\frac{3x-7}{2x-3}\)

⇒ 2x² – 9 + 3x = 3x² – 14 –x 

⇒ x² – 4x – 5 = 0

⇒ x² – 5x + x – 5 = 0 

⇒ (x – 5)(x + 1) = 0 

⇒ x = -1, 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.

\(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\)

⇒ 4(x² + 3x + (x – 1)(x – 2)) = 17(x² – 2x) 

⇒ 4(x² + 3x + x² – 3x + 2) = 17(x² – 2x) 

⇒ 8x² + 8 = 17x² – 34x 

⇒ 9x² – 34x – 8 = 0 

⇒ 9x2 – 36x + 2x – 8 = 0 

⇒ 9x(x – 4) + 2(x – 4) = 0 

⇒ (9x + 2)(x – 4) = 0 

⇒ x = -2/9, 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.

\(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x},x\neq0\)

⇒ x(x – 1 + 2x – 4) = 6(x² -3x + 2) 

⇒ 3x² - 5x = 6x² – 18x + 12 

⇒ 3x² - 13x + 12 = 0 

⇒ 3x² - 9x - 4x + 12 = 0 

⇒ 3x(x - 3) -4(x – 3) = 0 

⇒ (3x – 4)(x – 3) = 0 

⇒ x = 4/3, 3

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.

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

⇒ 6((x + 1)² - (x – 1)²) = 5(x² – 1) 

⇒ 6 × 4x = 5x² – 5 

⇒ 5x² – 24x – 5 = 0 

⇒ 5x² – 25x + x – 5 = 0 

⇒ 5x(x – 5) + 1(x-5) = 0 

⇒ (5x + 1)(x – 5) = 0 

⇒ x = 5, -1/5

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.

\(\frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2}, x\neq-\frac{1}{2},1\)

⇒ 2(x² - 2x + 1 + 4x² + 4x + 1) = 5(2x² – x – 1) 

⇒ 10x² + 4x + 4 = 10x² – 5x – 5 

⇒ 9x = -9 

⇒ x = -1

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.

\(\frac{16}{x}-1=\frac{15}{x+1};x\neq0,-1\)

⇒ (16 – x)(x + 1) = 15x 

⇒ -x² – x + 16x + 16 = 15x 

⇒ x² = 16

⇒ x = 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.

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

⇒ ax² +a²x + x + a =0 

⇒ ax(x + a) + 1(x + a) = 0 

⇒ (ax + 1)(x + a) = 0 

⇒ x = -a, -1/a

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.

\(3x^{2}-14x-5=0\)

⇒ 3x² – 15x + x – 5 = 0 

⇒ 3x(x – 5) + 1(x – 5) = 0 

⇒ (3x + 1)(x – 5) = 0 

⇒ x = 5, -1/3

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.

\(\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3};x\neq5,7\)

⇒ 3(x² – 11x + 28) + 3(x² – 11x + 30) = 10(x² – 12x + 35)

⇒ 4x² – 54x + 176 = 0 

⇒ 2x² – 27x + 88 = 0 

⇒ 2x² – 16x – 11x + 88 = 0 

⇒ 2x(x – 8) – 11(x – 8) = 0 

⇒ (2x – 11)(x – 8) = 0 

⇒ x = 11/2, 8

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.

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

⇒ x² – a² – x – a = 0 

⇒ (x + a)(x – a) – 1(x + a) = 0 

⇒ (x + a)(x – a – 1) = 0 

⇒ x = -a, a + 1

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.

\(\frac{4}{x}-3=\frac{5}{2x+3},x\neq0,-\frac{3}{2}\)

⇒ (4 – 3x)(2x + 3) = 5x 

⇒ 8x + 12 – 6x² – 9x = 5x 

⇒ 6x² + 6x – 12 = 0 

⇒ x² + x – 2 = 0 

⇒ x² + 2x – x – 2 = 0 

⇒ x(x + 2) – (x + 2) = 0 

⇒ (x – 1)(x + 2) = 0 

⇒ x = 1, - 2

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.

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

⇒ ax² + a – a²x – x = 0 

⇒ ax² – x(a² + 1) + a = 0 

⇒ ax(x – a) – 1(x – a) = 0 

⇒ (ax – 1)(x – a) = 0 

⇒ x = 1/a, a

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.

\((a+b)^{2}x^{2}-4abx-(a-b)^{2}=0\)

⇒ (a + b)²x² - (a² + b² + 2ab – a² – b² + 2ab)x – (a – b)² = 0 

⇒ (a + b)²x² – (a + b)²x + (a – b)²x – (a – b)² = 0 

⇒ (a + b)²x(x - 1) + (a – b)²(x – 1) = 0 

⇒ ((a + b)²x + (a – b)²)(x – 1) = 0

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

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.

\(x^{2}+2ab=(2a+b)x\)

⇒ x² – (2a + b)x + 2ab = 0 

⇒ x² – 2ax – bx + 2ab = 0 

⇒ x(x – 2a) – b(x – 2a) = 0 

⇒ (x – b)(x – 2a) = 0 

⇒ x = b, 2a

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.

\(7x+\frac{3}{x}=35\frac{3}{5}\)

⇒ 7x² + 3 = 178x/5 

⇒ 35x² – 178x + 15 = 0 

⇒ 35x² – 175x – 3x + 15 = 0 

⇒ 35x(x – 5) -3(x – 5) = 0 

⇒ (35x – 3)(x – 5) = 0 

⇒ x = 5, 3/35

Let the numbers be ‘a’ and ‘b’ 

Given, sum of two numbers is 16. The sum of their reciprocals is 1/3. 

⇒ a + b = 16 

⇒ a = 16 – b 

Also, 1/a + 1/b = 1/3 

⇒ 1/(16 – b) + 1/b = 1/3 

⇒ 3(b + 16 – b) = 16b - b² 

⇒ b² – 16b + 48 = 0 

⇒ b² – 12b – 4b + 48 = 0 

⇒ b(b – 12) – 4(b – 12) = 0 

⇒ (b – 4)(b – 12) = 0 

⇒ b = 4, 12 

Numbers are 4 , 12

Let the number be x.

Then from the question, we have

x + \(\frac{1}{x}\) = \(\frac{17}{4}\)

\(\frac{x^2 + 1}{x}\) = \(\frac{17}{4}\)

⇒ 4(x²+1) = 17x

⇒ 4x² + 4 – 17x = 0

⇒ 4x² + 4 – 16x – x = 0

⇒ 4x(x – 4) – 1(x – 4) = 0

⇒ (4x – 1)(x – 4) = 0

Now, either x – 4 = 0 ⇒ x = 4

Or, 4x – 1 = 0 ⇒ x = \(\frac{1}{4}\)

Thus, the value of x is 4.

Let the numbers be x and y (x > y).

According to the given condition,

x² – y² = 120 …(i) 

y² = 2x …(ii)

Substituting y² = 2x in equation (i), we get

x² – 2x = 120 

∴ x² – 2x – 120 = 0 

∴ x² – 12x + 10x – 120 = 0 

∴ x(x – 12) + 10(x – 12) = 0 

∴ (x – 12)(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 – 12 = 0 or x + 10 = 0 

∴ x = 12 or x = -10

But x ≠ -10

as, y² = 2x = 2(-10) = -20 …[Since, the square of number cannot be negative]

∴ x = 12

Smaller number = y² =2x

∴ y² = 2 × 12

∴ y² = 24

∴ y = ± √24 …[Taking square root of both sides]

∴ The smaller number is √24 and greater number is 12 or the smaller number is – √24 and greater number is 12.

Let the consecutive multiples of 3 be a, a + 3 

⇒ a(a + 3) = 270 

⇒ a² + 3a – 270 = 0 

⇒ a² + 18a – 15a – 270 = 0 

⇒ a(a + 18) -15(a + 18) = 0 

⇒ (a – 15)(a + 18) = 0 

⇒ a = 15 

Numbers are 15, 18

Let the two consecutive multiples of 3 be 3x and 3x + 3

From the question, it’s given that

3x*(3x + 3) = 270

⇒ x(3x + 3) = 90 [Dividing by 3 both sides]

⇒ 3x² + 3x = 90

⇒ 3x² + 3x – 90 = 0

⇒ x² + x – 30 = 0

Solving for x by factorization method, we have

x² + 6x – 5x – 30 = 0

⇒ x(x + 6) – 5(x + 6) = 0

⇒ (x + 6)(x – 5) = 0

Now, either x + 6 = 0 ⇒ x = – 6

Or, x – 5 = 0 ⇒ x = 5

Considering the positive value of x, we have only

x = 5, so 3x = 15 and 3x + 3 = 18.

Thus, the two consecutive multiples of 3 are 15 and 18 respectively.

Let one of the number be x. Then the other number will be 2x – 3.

From the question:

x² + (2x – 3)² = 233

⇒ x² + 4x² + 9 – 12x = 233

⇒ 5x² – 12x – 224 = 0

⇒ 5x² – 40x + 28x – 224 = 0

⇒ 5x(x – 8) + 28(x – 8) = 0

⇒ (5x + 28) (x – 8) = 0

Now, 5x + 28 cannot be 0

so, x – 8 = 0 ⇒ x = 8

Considering the value of x = 8, we have

2x – 3 = 15

Thus, the two numbers are 8 and 15 respectively.

Given that the product of two successive integral multiples of 5 is 300

Let’s assume the integers be 5x and 5(x+1), where x and x+1 are two consecutive multiples

Then, according to the question, we have

5x[5(x + 1)] = 300

⇒ 25x(x + 1) = 300

⇒ x² + x = 12

⇒ x² + x – 12 = 0

Solving for x by factorization method, we have

⇒ x² + 4x – 3x – 12 = 0

⇒ x(x + 4) – 3(x + 4) = 0

⇒ (x + 4)(x – 3) = 0

Now, either x + 4 = 0 ⇒ x = -4

Or, x – 3 = 0 ⇒ x = 3

For, x = – 4

5x = – 20 and 5(x + 1) = -15

And, for x = 3

5x = 15 and 5(x + 1) = 20

Thus, the two successive integral multiples can be 15, 20 or -15 and -20 respectively.

Let the number be ‘a’ 

⇒ a + a² = 63/4 

⇒ 4a² + 4a – 63 = 0 

⇒ 4a² + 18a – 14a – 63 = 0 

⇒ 2a(2a + 9) – 7(2a + 9) = 0 

⇒ (2a – 7)(2a + 9) = 0 

⇒ a = 7/2 or – 9/2

(d) 40 cm

Let the length of the hypotenuse be x cm. 

Then, length of base = (x – 2) cm 

x - 2 length of altitude =1

⇒ Length of altitude = \(\frac12(x - 1)\) cm

Applying Pythagoras' Theorem,

(Hyp.)² = (Base)² + (Perp.)²

⇒ x² = (x - 2)² + \(\big(\frac12(x-1)\big)^2\) 

⇒ x² = x² -4x + 4 + \(\frac14\)(x² - 2x + 1)

⇒ 4x² = 4(x² - 4x + 4) + (x² - 2x + 1)

⇒ 4x² = 4x² - 16x + 16 + x² - 2x +1

x² - 18x + 17 = 0

⇒ (x - 17)(x - 1) = 0 ⇒ x = 17, 1

x = 1 is not possible. 

∴ Length of hypotenuse = 17 cm 

Length of base = 15 cm

Length of altitude = \(\frac12\) x 16 cm = 8 cm

∴ Perimeter of the triangle = 17 cm + 15 cm + 8 cm = 40 cm.

(c) 44

x(x + 1) = 1980

⇒ x² + x - 1980 = 0

⇒ x² + 45x - 44x - 1980 = 0

⇒ x(x + 45) - 44(x + 45) = 0

⇒ (x + 45)(x - 44) = 0

⇒ x = - 45, 44 

Rejecting –45 , x = 44.