Explicação passo-a-passo:
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A)
\begin{gathered} \frac{x+x^2}{2}=210 \\ \\ x+x^2=2.(210) \\ \\ x+ x^{2} =420 \\ \\ x^{2} +x-420=0\end{gathered}
2
x+x
2
=210
x+x
2
=2.(210)
x+x
2
=420
x
2
+x−420=0
a=1
b=1
c=-420
Δ=b²-4ac
Δ=1²-4(1)(-420)
Δ=1+1680
Δ=1681
\begin{gathered}x= \frac{-b\pm \sqrt{\triangle} }{2a} = \frac{-1\pm41}{2} = \\ \\ x'= \frac{-1+41}{2} = \frac{40}{2} =20 \\ \\ x"= \frac{-1-41}{2} =- \frac{42}{2} =-21 \\ \\ S=\{-21,20\}\end{gathered}
x=
2a
−b±
△
=
2
−1±41
=
x
′
=
2
−1+41
=
2
40
=20
x"=
2
−1−41
=−
2
42
=−21
S={−21,20}
B)
\begin{gathered} x^{2} + \frac{3x}{5} =28 \\ \\ mmc=5 \\ \\ 5x^2+3x=140 \\ \\ 5x^2+3x-140=0\end{gathered}
x
2
+
5
3x
=28
mmc=5
5x
2
+3x=140
5x
2
+3x−140=0
a=5
b=3
c=-140
Δ=3²-4(5)(-140)
Δ=9+2800
Δ=2809
\begin{gathered}x= \frac{-3\pm53}{10} \\ \\ x'= \frac{-3+53}{10} = \frac{50}{10} =5 \\ \\ x"= \frac{-3-53}{10} =-\frac{56}{10} =-5,6~~ ou~~simplificando~~- \frac{28}{5} \end{gathered}
x=
10
−3±53
x
′
=
10
−3+53
=
10
50
=5
x"=
10
−3−53
=−
10
56
=−5,6 ou simplificando −
5
