Question

If
${x}^{2} + \frac{1}{ {x}^{2}} = 14$
Find:
$(1) \: x + \frac{1}{x}$
$(2) \: x - \frac{1}{x}$
$(3) \: {x}^{2} - \frac{1}{ {x}^{2} }$
​

Answer 1

Answer:

(i):   4         ;   (ii):    2√2         (iii):  8√2

Step-by-step explanation:

⇒ x² + 1/x² = 14

      Adding 2 to both sides:

⇒ x² + 1/x² + 2 = 14 + 2

⇒ x² + 1/x² + 2(1) = 16

⇒ x² + 1/x² + 2(x * 1/x) = 16

⇒ (x + 1/x)² = 16

⇒ x + 1/x = 4

Note: (x + 1/x)² = x² + 1/x² + 2   or

    (x + 1/x)² - 2 = x² + 1/x²

(ii):   x² + 1/x² = 14

         Adding -2 to both sides:

⇒ x² + 1/x² - 2 = 14 - 2

⇒ (x - 1/x)² = 12

⇒ x - 1/x = √12 = √(2²*3) = 2√3

  Note:  (x - 1/x)² = x² + 1/x² - 2   or

    (x - 1/x)² + 2 = x² + 1/x²

(iii):  x² - 1/x²

⇒ x² - (1/x)²

⇒ (x + 1/x)(x - 1/x)        [a² - b² = (a + b)(a - b)]

⇒ 4 * 2√2

⇒ 8√2

Answer 2

QuEsTiOn,

• ${x}^{2} + \frac{1}{ {x}^{2}} = 14$

To FiNd,

• $(1) \: x + \frac{1}{x}$
• $(2) \: x - \frac{1}{x}$
• $(3) \: {x}^{2} - \frac{1}{ {x}^{2} }$

SoLuTiOn,

(i): 4

x² + 1/x² = 14

Adding 2 to both sides:

x² + 1/x² + 2 = 14 + 2

x² + 1/x² + 2(1) = 16

x² + 1/x² + 2(x * 1/x) = 16

(x + 1/x)² = 16

⇒ x + 1/x = 4

Note:

• (x + 1 / x) ^ 2 = x ^ 2 + 1 / (x ^ 2) + 2

or

• (x + 1 / x) ^ 2 - 2 = x ^ 2 + 1 / (x ^ 2)

(ii): x ^ 2 + 1 / (x ^ 2) = 14

Adding -2 to both sides:

x² +1/x² - 2 = 14 - 2

(x - 1/x)² = 12

x - 1/x = √12 = √(2²*3) = 2√3

Note: (x - 1 / x) ^ 2 = x ^ 2 + 1 / (x ^ 2) - 2 or

(x - 1 / x) ^ 2 + 2 = x ^ 2 + 1 / (x ^ 2)

(iii): x² - 1/x²

x² - (1/x)²

→ (x + 1/x)(x - 1/x)

4* 2√2

→ 8√2

• [a ^ 2 - b ^ 2 = (a + b)(a - b)]

FiNaL AnSwEr,

1. 4
2. 2√2
3. 8√2

❥Hope you get your AnSwEr.

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