Question 56

If $$12x^2 - 21x + 1 = 0, $$ then what is the value of $$9x^2 + (16x^2)^{-1}$$?

Solution

$$12x^2 - 21x + 1 = 0 $$

Divide by 4x,

$$3x - 21/4 + 1/4x = 0 $$

$$3x + \frac{1}{4x} = \frac{21}{4}$$

On taking square,

$$(3x + \frac{1}{4x})^2 = (\frac{21}{4})^2$$

$$9x^2 + \frac{1}{16x^2} + 2.3x.\frac{1}{4x} = \frac{441}{16}$$

$$9x^2 + \frac{1}{16x^2} = \frac{441}{16} - \frac{3}{2}$$

$$9x^2 + \frac{1}{16x^2} = \frac{417}{16}$$

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