Question

Resolva as equações do 2° grau em R:

A) (2x-3)²=25
B) (x+1)²-x=7

Answer 1

Resolvendo por Bháskara:

$x = \frac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

$a) \\ (2x - 3)^2 = 25 \\ \\ (2x - 3) (2x - 3) = 25 \\ \\ 4x^2 -6x -6x + 9 - 25 = 0 \\ \\ 4x^2 - 12x-16 = 0$

a=4, b=−12, c=−16

Δ=b2−4ac
Δ=(−12)2−4*(4)*(−16)
Δ=144+256
Δ=400

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-12) \pm \sqrt{400}}{2*4} \\ \\ x = \frac{12 \pm 20}{8} \\ \\ x' = \frac{12 + 20}{8} \\ \\ x' = \frac{32}{8} \\ \\ x' = 4 \\ \\ \\ x'' = \frac{12 - 20}{8} \\ \\ x'' = \frac{-8}{8} \\ \\ x'' = -1$

S = {4, -1}
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$b) \\ (x + 1)^2 - x = 7 \\ \\ (x + 1) (x + 1) - x = 7 \\ \\ x^2 + x + x + 1 - x - 7 = 0 \\ \\ x^2 +x - 6 = 0$

a=1, b=1, c=−6

Δ=b2−4ac
Δ=(1)2−4*(1)*(−6)
Δ=1+24
Δ=25

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-1 \pm \sqrt{25}}{2*1} \\ \\ x = \frac{-1 \pm 5}{2} \\ \\ x' = \frac{-1 + 5}{2} \\ \\ x' = \frac{4}{2} \\ \\ x' = 2 \\ \\ \\ x'' = \frac{-1 - 5}{2} \\ \\ x'' = \frac{-6}{2} \\ \\ x'' = -3$

S = {2, -3}