Equations

1.
8x⁶ ‒9x³ + 1 = 0
8(x³)² ‒ 9(x³) + 1= 0
(x³ ‒ 1)(8x³ ‒1) = 0
(x³ ‒ 1)[(2x)³ ‒1)] = 0
(x ‒ 1)(x² + x + 1)(2x ‒ 1)[(2x)² +2x + 1] = 0
(x ‒ 1)(x² + x + 1)(2x ‒ 1)(4x² +2x + 1) = 0
x = 1 or x = (‒1 ± i√3)/2 or x= 1/2 or x = (‒1 ± i√3)/4

[If you have not learned about imaginary numbers,]
[the roots x = (‒1 ± i√3)/2 and x = (‒1 ± i√3)/4 are rejected.]


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2.
8x⁶ ‒10x⁴ ‒ 12x² =0
4x⁶ ‒ 5x⁴ ‒ 6x² = 0
x²(4x⁴ ‒ 5x² ‒ 6) =0
x²(4x⁴ ‒ 5x² ‒ 6) =0
x²(x² ‒ 2)(4x² +3) = 0
x = 0 (double roots) or x = ±√2 or x = ±i(√3)/2

[If you have not learned about imaginary numbers,]
[the roots x = ±i(√3)/2 are rejected.]


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3.
let y = √(x + 1)


2√(x + 1) ‒ 3 = 2 / √(x + 1)
2y ‒ 3 = 2/y
(2y ‒ 3)y = (2/y)y
2y² ‒ 3y = 2
2y² ‒ 3y ‒ 2 = 0
(y ‒ 2)(2y + 1) = 0
y = 2 or y = ‒1/2
√(x + 1) = 2 or √(x + 1) = ‒1/2 (rejected)
x + 1 = 4
x = 3


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4.
Let n and (n + 1) be the two consecutive integers.

[1 / n] + [1 / (n + 1)] = 5/6
[(n + 1) / n(n + 1)] + [n / n(n + 1)] = 5/6
[(n + 1) + n] / n(n + 1) = 5/6
(2n + 1) / (n² + n) = 5/6
5(n² + n) = 6(2n + 1)
5n² + 5n = 12n + 6
5n² ‒ 7n ‒ 6 = 0
(n ‒ 2)(5n + 3) = 0
n = 2 or n = ‒3/5 (rejected)
n + 1 = 3

Ans: The two consecutive integers are2 and 3.

1.

8x⁶ - 9x³ + 1 = 0
Let u = x³, then
8u² - 9u + 1 = 0
(8u - 1)(u - 1) = 0
u = 1/8 or 1
⇒ x³ = 1/8 or 1
x³ = (1/2)³ or 1³

⇒ x = 1/2 or 1

2.

8x⁶ - 10x⁴ - 12x² = 0
Let u = x², then
8u³ - 10u² - 12u = 0
(2u)(4u² - 5u - 6) = 0
(2u)(4u + 3)(u - 2) = 0
u = 0, -3/4 or 2
⇒ x² = 0, -3/4 or 2
x² = 0, x² + 3/4 = 0 (rej. since Δ < 0) or x² = 2

⇒ x = 0, √2 or -√2

3.

2√(x+1) - 3 = 2/√(x+1)
Let u = √(x+1), then
2u - 3 = 2/u
u(2u - 3) = 2
2u² - 3u - 2 = 0
(2u + 1)(u - 2) = 0
u = -1/2 or 2
⇒ √(x+1) = -1/2 (rej. since the value of a square root cannot be -ve) or 2
⇒ √(x+1) = 2
x+1 = 2² = 4

⇒ x = 3