Question

please solve the followings:- i) if x-1/x =root 5,find the value of x2+1/x2. ii) if x2+1/4x2=8,find x3+1/8x3. proper answer will be marked brainliest and irrelevant answers will be reported. plz answer it .​

Answer 1

Answer:

1). $\huge{\mathcal{\underline{\red{AnSWer}}}}$

GIVEN :

x-1/x=√5

Squaring both sides

(x-1/x)²=(√5)²

x²+1/x²-2×x×1/x=5

x²+1/x²=7

2). $\huge{\mathcal{\underline{\red{AnSWer}}}}$

We have x² + 1/(4x²) = 8.

But x² + 1/(4x²) = (x + 1/2x)² - 2×x×1/2x => (x + 1/2x)² -1 = 8

=> (x + 1/2x)² = 9

=> x + 1/2x = ±3

In both cases obtained value of x is real.

Now let S = x³ + 1/8x³ = (x + 1/2x)³ - 3×(x)×(1/2x)×(x+1/2x) = (x + 1/2x)³ - (3/2)×(x+1/2x)

S = (+3)³ - (3/2)(+3) or S = (-3)³ - (3/2)(-3)

=> S = 27 - 9/2 or -27+ 9/2

=> S = ± 22.5

So x³ + 1/8x³ = ± 22.5

Answer 2

Answer:

^2=power 2

v=square√

Step-by-step explanation:

x^2+1/x^2+2=9

(x+1/x)^2=9

x+1/x=3

tried my best plss mark as brainlest