Question

urgenteeee, bhaskara

Answer 1

Resposta:

$x^{2} -13x +36=0$

a) a= 1 b= -13 c= 36

b)Δ=b²-4ac

Δ= $(-13)^{2}$-4*1*36

Δ= 169-144

Δ=25

c)$x=\frac{-b+-\sqrt{Δ} }{2a}$

$x=\frac{-(-13)\sqrt{25} }{2*1}$

$x= \frac{13+-5}{2}$

$x'=\frac{13+5}{2} =\frac{18}{2} =9$

$x"=\frac{13-5}{2} =\frac{8}{2} =4$

d)

$y=(x')^{0,5} +(x")^{0,5}\\y=(9)^{0,5} +(4)^{0,5} \\y=3+2\\y=5$

Espero ter ajudado :)

Answer 2

Resposta:

$x^{2} -13x+36=0$

$a)\\ \\ coeficientes\\ \\ a=1\\ b=-13\\ c=36$

$b)\\ \\ discriminante~da~equacao~delta\\ \\ delta=b^{2} -4.a.c=(-13)^{2} -4.1.36=169-144~~~~delta=25$

$c)\\ \\ Bhaskara:\\ \\ x=\frac{-b+-\sqrt{delta} }{2.a} \\ \\ x=\frac{13+-\sqrt{25} }{2.1} \\ \\x=\frac{13+-5}{2} \\ \\ x'=\frac{13-5}{2}=\frac{8}{2} =4 \\ \\ x''=\frac{13+5}{2} =\frac{18}{2}=9$

$S~~\left \{ {{4;~9}} \}}$

$d)\\ \\ y=(x')^{0,5}+(x'')^{0,5}\\ \\ 0,5=\frac{1}{2} \\ \\ y=(x')^{\frac{1}{2} } +(x'')^{\frac{1}{2}$

$y=\sqrt{x'}+\sqrt{x''} =\sqrt{4} +\sqrt{9} =2+3=5\\ \\ y=5$