Question

find the set of each quadratic equation for y^2-8y+1 = 0


PLEASE ITS URGENT . PLEASE ​

Answer 1

Answer:

Given,

$y {}^{2} - 8y + 1 = 0$

Here, a = 1 ,b = -8 ,c = 1

By formula,

$y = \frac{ - b \frac{ + }{} \sqrt{b {}^{2} - 4ac} }{2a}$

$\: \: \: = \frac{ - ( - 8 )\frac{ + }{} \sqrt{( - 8) {}^{2} - 4(1)(1) } }{2(1)}$

$\: \: \: = \frac{8 \frac{ + }{} \sqrt{64 - 4} }{2}$

$\: \: \: = \frac{8 \frac{ + }{} \sqrt{60} }{2}$

$\: \: \: = \frac{8 \frac{ + }{} 2 \sqrt{15} }{2}$

$\: \: \: = \frac{2(4 \frac{ + }{} \sqrt{15} )}{2}$

$\: \: \: = 4 \frac{ + }{} \sqrt{15}$

$Thus , the \: solution \: set \: is \: \: ({4 + \sqrt{15} , \: 4 - \sqrt{15} })$