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Page 1: Math homework help service

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Math Homework Help Service

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About Us:

At Homework1.com we offer authentic and 100% accurate

online math homework help and study assistance to students

from USA, UK, Australia, and Canada. However, we don’t offer

students only math assignment help service to complete their

study project; rather we offer our best effort to teach our

student-clients about the homework we have solved. Our

tutors are not only subject matter experts, they are avid

student-mentors and are ready to walk extra miles to make them understand the

fundamentals of the assignment done, and help them to learn the solution by heart.

We are available online by 24×7 and we can be reached via email, live chat, as well as

by direct phone calls. Our USP is quick turnaround time with stringent quality assurance

about the assignment we undertake. We offer assistance is writing dissertations,

academic and project related essay writing, and in writing and reviewing research

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offer 100% plagiarism free content by following proper and prescribed house style.

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can first shortlist the reason you are looking for hiring this academic online support. This

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Homework Help at most reasonable cost.

Sample of Math Homework Help Illustrations and Solutions:

Illustration 1. Solve for x : 𝑥−1

𝑥−2 +

𝑥−3

𝑥−4 =

10

3 = (x ≠ 2, x ≠ 4)

Solution.

We have : 𝑥−1

𝑥−2 +

𝑥−3

𝑥−4 =

10

3 =

𝑥−1 𝑥−4 𝑥−2 (𝑥−3)

𝑥−2 𝑥−4 =

10

3

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= 𝑥2− 5𝑥+4 + (𝑥2−5𝑥+6)

𝑥2−6𝑥+8 =

10

3 =

2𝑥2− 10𝑥+10

𝑥2−6𝑥+8 =

10

3

= 10 𝑥2 - 10x – 5x + 25 = 0 = 2x (x -5) -5 (x – 5 ) = 0 = 2𝑥2 - 15x + 25 = 0

= 2𝑥2 - 10x – 5x + 25 = 0 = 2x (x – 5) -5 (x-5) = 0

= (x – 5 ) (2x – 5 ) = 0 = Either x – 5 = 0 or 2x -5 = 0 =x = 5 or x = 5

2

Hence, the solution are 5 and 5

2

Illustration 2. Solve the following quadratic equations by factorization method:

(i)4

𝑥 - 3 =

5

2𝑥+3 ; x ≠ 0,

3

2 (ii)

2𝑥

𝑥−3 +

1

2𝑥+3 +

3𝑥+9

𝑥−3 (2𝑥+3)= 0

Solution.

(i) We have 4

𝑥 - 3 =

5

2𝑥+3 =

4−3𝑥

𝑥 =

5

2𝑥+3

= (4 – 3x) (2x + 3) = 5x = 12 –x - 6𝑥2 = 5x = 6x2 + 6x – 12 = 0 = 𝑥2 +x – 2 = 0

= 𝑥2 + 2x – x – 2 = 0 = x (x +2) = 0 = (x+2) (x-1) = 0

= x + 2 = 0 or x – 1 = 0 = x = -2 or x = 1

(ii) Clearly, the given equation is valid if x – 3 ≠ 0 i.e., when x ≠- 3

2 , 3

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Now, 2𝑥

𝑥−3 +

1

2𝑥+3 +

3𝑥+9

𝑥−3 (2𝑥+3) = 0

= 2x (2x + 3) + (x-3) + 3x + 9 = 0 [Multiplying throughout by (x-3) (2x +

3)]

= 4𝑥2 + 6x +x -3 + 3x + 9 = 0 = 4𝑥2 + 10x + 6 = 0

= 2x2 + 5x + 3 = 0 = 2𝑥2 +2x + 3x + 3 = 0

= 2x (x + 1) + (3x + 1) = 0 = (2x + 3) (x + 1) = 0 = x + 1 = 0 = x = -1 [∵ 2x + 3 ≠

0]

Hence, x = -1 is the only solution of the given equation.

Illustration 3. The roots of the equation 𝑥2 - 3x + 2 = 0 are

(a) (1,-2) (b) (-1,2) (c) (-1,-2) (d) (-1,-2)

Solution.

x2 - 3x + 2 = 0 = (x – 1) (x – 2 ) = 0 = x = 1,2 = ∴ (d) holds.

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Illustration 4. The non-zero root of 3x - 5𝑥2 = 0 is

(a) 3

5 (b)

5

3 (c)

5

9 (d) (1,2)

Solution.

3x - 5𝑥2 = 0 = x (3 – 5x ) = 0 = x = 0 or x = 3

5

But x ≠ 0 ∴ x = 3

5 ∴ (a) holds.

Illustration 5. Which of the following equation has 2 as a root?

(a) 𝑥2 - 4x + 5 = 0 (b) 𝑥2 + 3x -12 = 0 (c) 2𝑥2 - 7x + 6 = 0 (d) 3𝑥2 - 7x + 6

= 0 (d)3𝑥2 - 6x-2= 0

Solution.

Clearly (c) holds [∵ 2 (2)2 -7 (2) + 6 = 8 – 14 + 6 = 0]

Illustration 6. The root of the quadratic equation 6𝑥2 -x -2 = 0 is :

(a) 1

2 (b)-

1

2 (c) -

2

3 (d) -1.

Solution.

6𝑥2 - x – 2 = 0 = 6𝑥2 – 4x + 3x – 2 = 0 = 2x (3x – 2) + 1 (3x – 2) = 0

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

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= 2x = -1 or 3x = 2 = x = −1

2 or x =

2

3 ∴ (b) holds.

Illustration 7. The positive root of 3𝑥2 + 6 = 9 is

(a) 3 (b) 4 (c) 5 (d) 7

Solution.

3𝑥2 + 6 = 9 = 3𝑥2 + 6 = 81 = 3𝑥2 = 81 – 6 = 75 = 𝑥2 = 25 = x = 5 ∴ (c) holds.

Illustration 8. Which of the following is a solution of the quadratic equation 𝑥2 - 𝑏2 = a

(2x – a) ?

(a) a + b (b) 2b – a (c) ab (d) 𝑎

𝑏

Solution.

We have 𝑥2- 𝑏2 = a (2x – a) = 2ax - 𝑎2= (𝑥2-2ax + 𝑎2) - 𝑏2 = 0 = (x – 𝑎)2 - 𝑏2 = 0

= (x – a – b) (x –a + b) = 0 = x –a –b = 0 or x – a + b = 0

= x = a + b or x = a –b

Illustration 9. The roots of the equation 𝑥2 + 5x – (𝛼 + 1 ) (𝛼 + 6) = 0,

where 𝛼 is a constant, are

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(a) 𝛼 + 1, 𝛼 + 6 (b) (𝛼 + 1) 𝛼 + 6 = 0

(c)- (𝛼 + 1), (𝛼 + 6) (d) – (𝛼 + 1), - (𝛼 + 6)

Solution.

The given equation can be written as

𝑥2 + [(a + 6) – (𝛼 + 1)] x – (𝛼 + 1) 𝛼 + 6 = 0

= x (x +(𝛼 + 6) - ( + 1)(x (𝛼 + 6) =0

= (x + (𝛼 + 6)) (x – (𝛼 + 1)) = 0 = x = -(𝛼 + 6), 𝛼 + 1. ∴ (c) Holds,