Question

F(x)=x^2+x-1,if f(x)=5,find the value of x

Answer 1

• Solving by Shree Dhracharya method :-

x² + x - 1

D = b² - 4ac

D = 1 - 4 × 1 × - 1

D = 1 + 4

D = 5


______

x = ( - b ± √D ) / 2a

x = ( - 1 ± √5 ) / 2a

$x = \frac{ - 1 + \sqrt{5} }{2}$

$x = \frac{ - 1 - \sqrt{5} }{2}$