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

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

Uniteasy Admin · Accepted Answer

$$\because   a^2−4a−1=0$$
$$	herefore  a  -\dfrac{1}{a} =4   $$n
Square both side, then
$$a^2+\dfrac{1}{a^2} = 18  $$n
$$a^2+3a+\dfrac{1}{a^2}-\dfrac{3}{a} $$
$$=a^2+\dfrac{1}{a^2}+3(a  -\dfrac{1}{a})$$
$$=18+3\cdot 4 =30$$
therefore 
D is the right choice

Multiple Choice Question (MCQ)

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