Question

If $(4y-\frac{4}{y})=11$ , find the value of $(y^2+\frac{1}{y^2})$.

• A. $7\frac{9}{16}$
• B. $5\frac{9}{16}$
• C. $9\frac{11}{16}$
• D. $9\frac{9}{16}$

Answer 1

Answer:

D. $9\frac{9}{16}$

Solution:

Given:

$4y-\frac{4}{y} = 11$

Formula used:

(a + b)²  = a² + 2ab + b²

Calculation:

$4y-\frac{4}{y} = 11$

Dividing the whole equation by 4.

⇒ $y^2+\frac{1}{y^2}=(\frac{11}{4})^2+2$

⇒ $\frac{121}{16} + 2$

⇒ $\frac{153}{16} = 9(\frac{9}{16})$

The answer is $9(\frac{9}{16})$