[tex]1+\dfrac{\sqrt{5}}{2},1-\dfrac{\sqrt{5}}{2}[/tex]
Step-by-step explanation:
The given equation is [tex]4x^{2}-8x-1[/tex]
Let [tex]a[/tex] be the coefficient of [tex]x^{2}[/tex]
Let [tex]b[/tex] be the coefficient of [tex]x[/tex]
Let [tex]c[/tex] be the constant.
Then the roots α,β for the equation [tex]ax^{2}+bx+c[/tex] are [tex]\dfrac{-b+\sqrt{b^{2}-4ac} }{2a},\dfrac{-b-\sqrt{b^{2}-4ac} }{2a}[/tex]
So,α=[tex]\frac{-b+\sqrt{b^{2}-4ac} }{2a}=\frac{8+\sqrt{64+16} }{8}=\frac{8+4\sqrt{5}}{8}=1+\frac{\sqrt{5}}{2}[/tex]
β=[tex]\frac{-b-\sqrt{b^{2}-4ac} }{2a}=\frac{8-\sqrt{64+16} }{8}=\frac{8-4\sqrt{5}}{8}=1-\frac{\sqrt{5}}{2}[/tex].
So the roots are [tex]1+\frac{\sqrt{5}}{2},1+\frac{\sqrt{5}}{2}[/tex]
