Question

17
28. Solve for x : x^2+1/x^2 = 17/4​

Answer 1

$x^2+\dfrac{1}{x^2}=\dfrac {17}{4}\\\\\\x^2+\dfrac {1}{x^2}+2=\dfrac {17}{4}+2\\\\\\\left (x+\dfrac {1}{x}\right)^2=\dfrac {25}{4}\\\\\\x+\dfrac {1}{x}=\pm\dfrac {5}{2}\\\\\\\dfrac {x^2+1}{x}=\pm\dfrac {5}{2}\\\\\\2x^2\pm 5x+2=0\\\\\\2x^2\pm 4x\pm x+2=0\\\\\\2x(x\pm 2)\pm(x\pm 2)=0\\\\\\(x\pm 2)(2x\pm 1)=0\\\\\\\implies\ \mathbf {x=\pm 2}\quad\text{OR}\quad \mathbf {x=\pm \dfrac {1}{2}}$

Answer 2

Step-by-step explanation:

x

2

+

x

2

1

=

4

17

x

2

+

x

2

1

+2=

4

17

+2

(x+

x

1

)

2

=

4

25

x+

x

1

=±

2

5

x

x

2

+1

=±

2

5

2x

2

±5x+2=0

2x

2

±4x±x+2=0

2x(x±2)±(x±2)=0

(x±2)(2x±1)=0

⟹ x=±2ORx=±

2

1