Question

x²=3-2x fórmula de Bhaskara ​

Answer 1

Olá!

x² = 3 - 2x

x² + 2x - 3 = 0

a = 1

b = 2

c = -3

Δ = b² - 4.a.c

Δ = 2² - 4. 1. (-3)

Δ = 4 + 12

Δ = 16

x = -b ±√Δ / 2.a

x = - (2) ± √16 / 2 . 1

x = -2 ± 4 / 2

x' = $\frac{-2-4}{2}$ = -3

x" = $\frac{-2+4}{2}$ = 1

S = {-3; 1}

Espero ter ajudado :)

Answer 2

As Raízes da equação são; S = {1,-3}

• Antes de começar temos que simplificar.
• Simplificando.

$\boxed{\begin{array}{lr} x^2=3-2x\\\boxed{x^2+2x-3=0\ \ \checkmark} \end{array}}$

• Para acharmos as Raízes da Equação, temos que ter seu coeficientes, que são eles.

A = 1

B = 2

C = -3

• Agora resolvendo o Delta mais conhecido como Discriminante.

$\boxed{\begin{array}{lr} \Delta=b^2-4.a.c\\\Delta=2^2-4.1.-3\\\Delta=4+12\\\Delta=16 \end{array}}$

• Agora basta resolver;

$\boxed{\begin{array}{lr} x=\dfrac{-2\pm4}{2} \end{array}}$

• Resolvendo ( x' e x'').

$\boxed{\begin{array}{lr} x'=\dfrac{-2+4}{2} \end{array}}\to\boxed{\begin{array}{lr} x'=\dfrac{2}{2} \end{array}}\to\boxed{\begin{array}{lr} \boxed{\begin{array}{lr} x'=1\ \ \checkmark \end{array}} \end{array}}\\\\\\\boxed{\begin{array}{lr} x''=\dfrac{-2-4}{2} \end{array}}\to\boxed{\begin{array}{lr} x''=\dfrac{-6}{2} \end{array}}\to\boxed{\begin{array}{lr} \boxed{\begin{array}{lr} x''=-3 \ \ \checkmark \end{array}} \end{array}}$

Resposta;

S = {1,-3}

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