If 2x2 - 7x + 5 = 0, then what is the value of $${x^2} + \frac{{25}}{{4{x^2}}}?$$
A. $$5\frac{1}{2}$$
B. $$7\frac{1}{4}$$
C. $$9\frac{1}{2}$$
D. $$9\frac{3}{4}$$
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & 2{x^2} - 7x + 5 = 0 \cr & 2x + \frac{5}{x} = 7 \cr & x + \frac{5}{{2x}} = \frac{7}{2} \cr & {\left( {x + \frac{5}{{2x}}} \right)^2} = {\left( {\frac{7}{2}} \right)^2} \cr & {x^2} + \frac{{25}}{{4{x^2}}} + 2 \times x \times \frac{5}{{2x}} = \frac{{49}}{4} \cr & {x^2} + \frac{{25}}{{4{x^2}}} = \frac{{49}}{4} - 5 \cr & {x^2} + \frac{{25}}{{4{x^2}}} = \frac{{29}}{4} \cr & {x^2} + \frac{{25}}{{4{x^2}}} = 7\frac{1}{4} \cr} $$Related Questions on Algebra
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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