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Homework1
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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
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study project; rather we offer our best effort to teach our
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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
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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,