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Question 75
If $$20x^{2} — 30x + 1 = 0$$, then what is the value of $$25x^{2}+\frac{1}{16x^{2}}$$
Solution
$$20x^2-30x+1=0$$
Dividing by x
20x-30+1/x=0
20x+1/x=30
5x+1/4x=15/2
$$25x^{2}+\frac{1}{16x^{2}}$$
=$$25x^2 + \frac{1}{16x^2} +2 \times 5x \times 1/4x - 2 \times 5x \times 1/4x$$
$$= 25x^2 + \frac{1}{16x^2} + \frac{5}{2} - \frac{5}{2}$$
$$=(5x+1/4x)^2 - \frac{5}{2}$$
= $$(\frac{15}{2}) - \frac{5}{2}$$
= $$\frac{225}{4} - \frac{5}{2}$$
= $$\frac{215}{4} = 53\frac{3}{4}$$
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