Question 123

If $$a^{2}-4a-1 = 0$$, then value of $$a^{2}+\frac{1}{a^{2}}+3a-\frac{3}{a}$$ is

Solution

it is given that $$a^{2}-4a-1 = 0$$

from this we can say a - $$\frac{1}{a}$$ = 4

we need to find $$a^{2}+\frac{1}{a^{2}}+3a-\frac{3}{a}$$

$$a^2 + \frac{1}{a^2}$$ = $$(a - \frac{1}{a})^2 + 2$$

$$a^{2}+\frac{1}{a^{2}}+3a-\frac{3}{a}$$ = $$4^2 + 2 + (3x4)$$

= 30

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