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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