Question

please anybody answer this very urjent i have a exam today​

Answer 1

${x}^{2} - \frac{4}{ {x}^{2} } = 3 \\ \\ {x}^{2} - \frac{4}{ {x}^{2} } - 3 = 0$

Taking LCM=x²

Terms Can be written as

$\frac{ {x}^{2} \times {x}^{2} - 4 - 3 \times {x}^{2} }{ {x}^{2} } = 0 \\ \\ \frac{ {x}^{4} - 4 - 3 {x}^{2} }{ {x}^{2} } = 0 \\ \\ {x}^{4} - 4 - {3x}^{2} = 0 \times {x}^{2} \\ \\ {x}^{4} - {3x}^{2} - 4 = 0 \\ \\ {x}^{4} - {4x}^{2} + {x}^{2} - 4 = 0 \\ \\ ( {x}^{4} - {4x}^{2} ) + ( {x}^{2} - 4) = 0 \\ \\ {x}^{2} ( {x}^{2} - 4) + 1( {x}^{2} - 4) = 0 \\ \\( {x}^{2} + 1)( {x}^{2} - 4) = 0$

Means

x²+1=0

x²=-1

x=√-1

x²-4=0

x²=4

x=√4

x=2

So,

X=2

CHECK

x²=4

x²-4/x²=3

4-4/4

16-4/4

12/4

3

Hence

LHS=RHS

X=2

Answer 2

Answer:

Step-by-step explanation: