Question

Find the roots of x²-3x-1 =0 by using the quadratic formula​

Answer 1

hope this will help you........

Answer 2

     

$p(x)=x^2-3x-1=0 \\ \\ \\ a=1 \\ \\ b=-3 \\ \\ c=-1$

$\leadsto\ b^2-4ac \\ \\ \leadsto\ (-3)^2-(4 \times 1 \times -1) \\ \\ \leadsto\ 9-(-4) \\ \\ \leadsto\ 9+4 \\ \\ \leadsto\ 13$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-(-3)\pm\sqrt{13}}{2 \times 1} \\ \\ x=\bold{\frac{3\pm\sqrt{13}}{2}} \\ \\ \\ \\ \therefore\ The\ roots\ of\ \ p(x)=x^2-3x-1=0\ \ are\ \ \bold{\frac{3+\sqrt{13}}{2}}\ \ \&\ \ \bold{\frac{3-\sqrt{13}}{2}}.$

$Thank\ you... \\ \\ \\ \mathfrak{\#adithyasajeevan}$