Question

2. Solve the following quadratic equations using quadratic formula

Answer 1

Given,

$x+\frac{1}{x}=2\frac{1}{2}\\\\\frac{x^{2}+1}{x}=\frac{5}{2}\\\\2x^{2}+2=5x\\\\2x^{2}-5x+2=0\\ \\2x^{2}-4x-x+2=0\\\\2x(x-2)-1(x-2)=0\\\\(x-2)(2x-1)=0\\\\Hence,\\x=2\;\;or\;\;x=\frac{1}{2}$

Answer 2

solution :-

given by:-
$x + \frac{1}{x} = 2 \frac{1}{2} \\ \frac{ {x}^{2} + 1 }{x} = \frac{5}{2} \\ 2{x}^{2} + 2 - 5x = 0 \\ 2 {x}^{2} - 5x + 2 = 0 \\a = 2 \\ \: b = - 5 \\ \: c = 2 \\ by \: formula \\ \frac{ - b( + - ) \sqrt{ {b}^{2} - 4ac } }{2a} \\ \frac{5( + - ) \sqrt{25 - 4 \times 2 \times 2} }{2 \times 2} \\ \frac{5( + - ) \sqrt{25 - 16} }{4} \\ \frac{5( + - ) \sqrt{9} }{4} \\ \frac{5( + - )3}{4} \\ x = \frac{8}{4} = 2 \\ x = \frac{2}{4} = \frac{1}{2}$
☆i hope its help☆