If ​​​​​ $$a²- 4a +1= 0$$, find $$a² + \frac{1}{a^2}$$.​

Correct option is A

Given:

$$a^2 - 4a + 1 = 0$$ 

$$a^2 + \frac{1}{a^2}$$ = ? 

Concept Used:

​$$(a+b)^2= a^2+b^2+2ab$$ 

Or  $$a^2+b^2= (a+b)^2-2ab$$​​

Solution:

solving the quadratic equation: 

$$a^2 - 4a + 1 = 0$$ 

both sides divided by 'a'

​$$\frac{1}{a} (a^2-4a+1)= \frac{0}{a}$$ 

$$a-4+\frac{1}{a}= 0$$ 

​$$a +\frac 1 a = 4$$​​

To find $$a^2 + \frac{1}{a^2}$$​ , we will use the identity:

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

​Now that we know $$a + \frac{1}{a} = 4$$, substituting this into the identity: 

​$$a^2 + \frac{1}{a^2} = 4^2 - 2 = 16 - 2 = 14$$ 

Thus the value of $$( a^2 + \frac{1}{a^2})$$ is 14.​

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