Question

if x² + 1/ x² = 223 and x > 0, find the value of the following:

a) ( x + 1/x)

b) ( x - 1/ x)²​​​

Answer 1

Step-by-step explanation:

Given :-

x² + 1/ x² = 223 and x > 0,

To find :-

Find the value of the following:

a) ( x + 1/x)

b) ( x - 1/ x)²

Solution :-

Given that

x² + 1/ x² = 223 -------(1)

and x > 0

It can be written as

=> (x)² +(1/x)² = 223

LHS is in the form of a²+b²

Where, a = x and b = 1/x

We know that

(a+b)² = a²+2ab+b²

=> a²+b² = (a+b)²-2ab

Now,

x²+(1/x)² = [x+(1/x)]²-2(x)(1/x)

=> x²+(1/x)² = [x+(1/x)]² -2(x/x)

=> x²+(1/x)² = [x+(1/x)]² -2(1)

=> x²+(1/x)² = [x+(1/x)]² -2

Now,

from (1)

=> [x+(1/x)]² -2 = 223

=> [x+(1/x)]² = 223+2

=> [x+(1/x)]² = 225 -------(2)

=> x+(1/x) = ±√225

=> x+(1/x) = ±15

Since ,x > 0

=> x+(1/x) = 15

And

We know that

(a-b)² = (a+b)²-4ab

[x-(1/x)]² = [x+(1/x)]²-4(x)(1/x)

=> [x-(1/x)]² = [x+(1/x)]²-4(x/x)

=> [x-(1/x)]² = [x+(1/x)]²-4(1)

=> [x-(1/x)]² = [x+(1/x)]²-4

=> [x-(1/x)]² = 225-4

From (2)

=> [x-(1/x)]² = 221

Answer:-

a) The value of x+(1/x) for the given problem is 15

b) The value of [x-(1/x)]² for the given problem is 221

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ (a-b)² = (a+b)²-4ab