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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
Homework1
Copyright © 2014-2015 Homework1.com, All rights reserved
=
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
Homework1
Copyright © 2014-2015 Homework1.com, All rights reserved
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.
Homework1
Copyright © 2014-2015 Homework1.com, All rights reserved
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
Homework1
Copyright © 2014-2015 Homework1.com, All rights reserved
= 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
Homework1
Copyright © 2014-2015 Homework1.com, All rights reserved
(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,

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Math Homework Help

  • 1. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved Math Homework Help Service Contact Us Homework1 3422 SW 15 Street Suite #8924 Deerfield Beach, FL, US 33442 Tel: +1-626-472-1732 Web: https://homework1.com/ Email: info@homework1.com Facebook: https://www.facebook.com/homework1com Linkedin: https://www.linkedin.com/in/homework1 Twitter: https://twitter.com/homework1_com Google Plus: https://plus.google.com/118210863993786098250/ Pinterest: https://www.pinterest.com/homeworkone/
  • 2. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved 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 papers. These research papers are done by best subject matter experts available and we offer 100% plagiarism free content by following proper and prescribed house style. Looking for quality math homework help ? Before you hire this academic service, you can first shortlist the reason you are looking for hiring this academic online support. This is not the fact you can only get to solve complicated math assignment problems, but you will be able to learn and understand the problem pretty well by some of the most expert tutors on the most interactive communication platform by hiring this service. In one word you can be benefited in a multifaceted way provided you can select the best service provider from the market. We at homework1 not only honor your requirement, we understand the necessity of doing the assignment by your deadline no matter you have contacted us the most critical hour. Besides maintaining best quality, we also offer latest updates in the solution so that you can impress your reviewer. We offer quality Math 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
  • 3. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved = 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
  • 4. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved 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.
  • 5. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved 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
  • 6. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved = 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
  • 7. Homework1 Copyright © 2014-2015 Homework1.com, All rights reserved (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,