Question

Resolva a seguinte equações do 2-0 grau x2 + 5x +6= 0 S={-3,-2}. S={-9,-5}. S={2,9}. S={3,2}

Answer 1

Resposta:

S={–3;–2}

Explicação passo-a-passo:

x²+5x +6=0

X= – b +/– √ b² – 4ac / 2a

x = – 5 +/– √ 5² – 4 × 1×6/2×1

x = –5 +/– √25 –24/2

x = –5+/–√1/2

x = –5+/–1/2

x,= –5–1/2

x, = –6/2

x, = –3„

x,,= –5+1/2

x,, = –4/2

x,,= –2„

Answer 2

Equação Do Segundo Grau

• Coeficientes:

$\boxed{ \begin{array}{lr} \large \sf a = 1 \\ \large \sf b = 5 \\ \large \sf c = 6 \end{array}}$

• Discriminante:

$\boxed{ \begin{array}{lr} \\ \large \sf \: \Delta = {b}^{2} - 4ac \\ \\ \: \large \sf \:\Delta = {(5)}^{2} - 4 \cdot 1\cdot6 \\ \\ \large \sf \:\Delta = 25 - 24 \\ \\ \red{\large \sf \: \Delta = 1} \\ \: \end{array}}$

• Bhaskara:

$\boxed{ \begin{array}{lr} \\ \large \sf \: x = \dfrac{ - b \: \pm \: \sqrt{\Delta} }{2.a} \\ \\ \\ \large \sf \: x = \dfrac{ - ( + 5) \: \pm \: \sqrt{1} }{2.1} \\ \\ \\ \large \sf \: x = \dfrac{ -5 \pm \: 1 }{2} \\ \: \end{array}}$

• Raízes:

$\large \boxed{ \boxed{ \large \sf \: x_{1} = \dfrac{ - 5 + 1}{2} = \boxed{ \red{\sf - 2}}}} \\ \\ \\ \large \boxed{ \boxed{ \large \sf \: x_{2} = \dfrac{ - 5 - 1}{2} = \boxed{ \red{\sf - 3}}}}$

➡️ Resposta:

$\bullet \: \huge \boxed{ \boxed{ \huge \sf \: \: S = \{ - 2, - 3\} \: \: }}$

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