Question

x is equal to 9 minus 4 under root 5 find the value of x square + 1 upon x square​

Answer 1

Given that x=9-4√5

So,

$\frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} }$

To make this easier, let us rationalize the denominator.

$\frac{1}{x} = \frac{1}{(9 - 4 \sqrt{5}) } \times \frac{(9 + 4 \sqrt{5} )}{(9 + 4 \sqrt{5} )}$

[(a+b) (a-b)=a²-b²]

$\frac{1}{x} = \frac{9 + 4 \sqrt{5} }{9 {}^{2} - (4 \sqrt{5} ) {}^{2} }$

$\frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - (16 \times 5 )}$

$\frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80}$

$\frac{1}{x} = \frac{9 + 4 \sqrt{5} }{1}$

$\frac{1}{x} = 9 + 4 \sqrt{5}$

So,

$x {}^{2} + \frac{1}{ {x}^{2} } = (9 - 4 \sqrt{5} ) {}^{2} + (9 + 4 \sqrt{5}) {}^{2}$

For (9-4√5)² use [ (a-b)²=a²+b²-2ab ]

For (9+4√5)² use [ (a+b)²=a²+b²+2ab ]

So,

$(9 {}^{2} +(4 \sqrt{5} ) {}^{2} - 2 \times 9 \times 4 \sqrt{5} ) + (9 {}^{2} + (4 \sqrt{5} ) {}^{2} + 2 \times 9 \times 4 \sqrt{5} )$

$161 - 72 \sqrt{5} + 161 + 72 \sqrt{5}$

-72√5 and 72√5 get cancelled.

So,

$161 + 161$

$322$