Question : If $2 x^2-7 x+5=0$, then what is the value of $x^2+\frac{25}{4 x^2} ?$
Option 1: $5 \frac{1}{2}$
Option 2: $7 \frac{1}{4}$
Option 3: $9 \frac{1}{2}$
Option 4: $9 \frac{3}{4}$
Correct Answer: $7 \frac{1}{4}$
Solution :
$2x^2 – 7x + 5 = 0$
$⇒2x^2 + 5 = 7x$
Dividing by $2x$, we get,
$⇒x + \frac{5}{2x} = \frac{7}{2}$
Squaring both sides, we get,
$⇒(x + \frac{5}{2x})^2 = (\frac{7}{2})^2$
$⇒x^2 + (\frac{5}{2x})^2 + 2 × x × \frac{5}{2x} = \frac{49}{4}$
$⇒x^2 + \frac{25}{4x^2} + 5 = \frac{49}{4 }$
$⇒x^2 + \frac{25}{4x^2} = \frac{49}{4} - 5$
$⇒x^2 + \frac{25}{4x^2} = \frac{29}{4}$
$\therefore x^2 + \frac{25}{4x^2} = 7\frac{1}{4}$
Hence, the correct answer is $7\frac{1}{4}$.
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