Question

if x=5-√3/5+√3 then find x^2-8x+1​

Answer 1

Answer:

Step-by-step explanation:

$x=\frac{5-\sqrt{3} }{5+\sqrt{3}} \\x=\frac{5-\sqrt{3} }{5+\sqrt{3}} *\frac{5-\sqrt{3} }{5-\sqrt{3}} \\x=\frac{(5-\sqrt{3})^{2} }{25-9}\\ x=\frac{25+9-10\sqrt{3} }{16} \\x=\frac{24-10\sqrt{3} }{16} =\frac{12-5\sqrt{3} }{8} =\frac{3}{2}-\frac{5}{8}\sqrt{3}$

In the quadratic equation, their is no way t factorize it ; so we will directly substitute x as the value given above.

$x^{2} -8x+1\\=(\frac{3}{2}-\frac{5}{8}\sqrt{3})^{2} +8(\frac{3}{2}-\frac{5}{8}\sqrt{3}) +1\\= \frac{9}{4} +\frac{25}{64} *3 -(2*\frac{3}{2} *\frac{5}{8}\sqrt{3} )+12-5\sqrt{3}$

On solving or simplifying :

$\frac{1051}{64} -\frac{55\sqrt{8} }{8}$ or $\frac{1}{64} (1051-440\sqrt{3} )$ or $4.51402569796396860699880640214712747726821388005363318211632701...$

Answer 2

Answer:

this is very difficult dont even try ur head will blast into peices bye bye take care