Question

4Xao quadrado-3X-1=0

Answer 1

$4x^2-3x-1=0 \\ \\ a=4 \\ b=-3 \\ c=-1 \\ \\ \\ x= \dfrac{-b\pm \sqrt{b^2-4\,a\,c} }{2\,a} \\ \\ \\ x = \dfrac{-(-3)\pm\sqrt{(-3)^2-4\cdot4\cdot(-1)}}{2\cdot4} \\ \\ \\ x = \dfrac{3\pm\sqrt{9-(-16)}}{ 8 } \\ \\ \\ x=\dfrac{3\pm \sqrt{25} }{8} \\ \\ \\ x = \dfrac{3\pm5}{8} \\ \\ \\ x'= \frac{3+5}{8} \rightarrow \boxed{x'=1} \\ \\ x''= \frac{3-5}{8} \rightarrow \boxed{x''=- \frac{1}{4} }$

Answer 2

4x²-3x-1=0

a=4 b=-3 c=-1

$x= \frac{ -b_{-}^+\sqrt{b^2- 4.a.c}}{2.a} = \frac{ -(-3)_{-}^+\sqrt{-3^2- 4.4.(-1)}}{2.4} =$

$\frac{ 3_{-}^+\sqrt{9+16}}{8} = \frac{ 3_{-}^+\sqrt{25}}{8} = \frac{ 3_{-}^+{5}}{8}$


$x^I= \frac{ 3+{5}}{8} = \frac{8}{8} = 1 \\ \\ x^I^I= \frac{ 3-{5}}{8} = \frac{-2}{8} = \frac{-1}{4}$