Question

y2+1/y2=14 . find y+1/y​

Answer 1

y^2+1/y^2=14

y^2+1/y^2+2y×1/y-2y×1/y=14 {add and subtract 2y×1/y}

(y+1/y)^2-2y×1/y=14. {(a+b)^2=a^2+b^2+2ab}

(y+1/y)^2-2=14

(y+1/y)^2=16

y+1/y=√16

y+1/y=+ or - 4

Answer 2

The value of the given expression of the sum of 'y' and its inverse is 4 or -4 alternatively.

The given equation of the sum of squares of "y" and its inverse is :

Equation 1 :

            $y^{2} + 1/y^{2} = 14$

We need to find the value of the other equation which is the sum of "y" and its inverse. Let us assume this required value to be "x".

Equation 2 :

             $y + 1/y = x$

       

Squaring both sides of Equation 2 above, we get :

              $y + 1/y = x\\(y + 1/y)^{2} = x^{2}$

Using the identity: $(a+b)^{2} = a^{2} + 2ab + b^{2}$ to open the square brackets,

             $(y)^{2} + 2(y)(1/y) + (1/y)^{2} = x^{2} \\(y)^{2} + (1/y)^{2} + 2= x^{2}$

Replacing the squared values from Equation 1 :

             $14 + 2 = x^{2} \\x^{2} = 16\\x = 4, -4$

Hence, the value of the given expression of 'y' and its inverse can be either 4 or -4.

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