Question

plzz solve this question if anyone knows...​

Answer 1

Integral Solution

• A solution which is an integer. (integer + -al)

So let's solve the problem.

We can change the form of $\sf{y^2+\frac{1}{y^2} }$ with a suitable identity.

On squaring $\sf{y+\frac{1}{y} }$

$\sf{(\sf{y+\frac{1}{y} })^2=y^2+2\times(y)\times(\frac{1}{y} )+(\frac{1}{y} )^2}$

$\sf{(\sf{y+\frac{1}{y} })^2=y^2+2+\frac{1}{y^2}$

Transposing the Sides

∴$\sf{(\sf{y+\frac{1}{y} })^2-2=y^2+\frac{1}{y^2}$

Now the equation becomes

$\sf{-2(y+\frac{1}{y})^2+4+7(y+\frac{1}{y})=9 }$

$\sf{-2(y+\frac{1}{y})^2+7(y+\frac{1}{y})-5=0 }$

$\sf{2(y+\frac{1}{y})^2-7(y+\frac{1}{y})+5=0 }$

Using the factorization method

$\sf{2t^2-7t+5=0}$

2t             -5

t               -1

The zeros are $\sf{y+\frac{1}{y}=\frac{5}{2}$ or $\sf{y+\frac{1}{y}=1}$

The two integral zeros come from the first equation.

$\sf\therefore{There \ are \ two \ integral \ solutions.$

Answer 2

Step-by-step explanation:

In this way u can solve.....

after solving u can get the answer