Question

${\displaystyle x^2-6x+4=0}$ 

 

Answer 1

Given: 

${\displaystyle x^2-6x+4=0}$ 

On comparing it with ${\displaystyle a{x}^2+bx+c=0}$ 

a =1,b = -6 and c = 4  

Discriminant D is given by: 

${\displaystyle D=(b^2-4ac)}$

=${\displaystyle {(-6)}^2-4\times 1\times 4}$ 

=${\displaystyle 36-16}$ 

=${\displaystyle 20>0}$  

Hence, the roots of the equation are real.
Roots α and β are given by: 

${\displaystyle \alpha =\frac{-b+\sqrt{D}}{2c}=\frac{-(-6)+\sqrt{20}}{2\times 1}=\frac{6+2\sqrt{5}}{2}=\frac{2(3+\sqrt{3})}{2}=(3+\sqrt{5})}$  

${\displaystyle \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-(-6)-\sqrt{20}}{2}=\frac{6-2\sqrt{5}}{2}=\frac{2(3-\sqrt{5})}{2}=(3-\sqrt{5})}$ 

Thus, the roots of the equation are ${\displaystyle (3-2\sqrt{5})\text{and}(3-2\sqrt{5})}$