If $$x + \frac{9}{x} = 6{\text{,}}$$ then $$\left( {{x^2} + \frac{9}{{{x^2}}}} \right)$$ is equal to?
A. 8
B. 9
C. 10
D. 12
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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