Question

x² - 2x = x + 4 me ajuda por favor ​

Answer 1

x¹ = -1

x² = 4

$\Large{\sf{\Huge ————————————————}}$

$\sf \red{ x^{2} } \green{ - } \blue{ 2x } \pink{ = } \purple{ x } \orange{ + } \red{ 4 }\\ \sf \green{ x^{2} } \blue{ - } \pink{ 2x } \purple{ - } \orange{ x } \red{ - } \green{ 4 } \blue{ = } \pink{ 0 } \\ \sf \purple{ x^{2} } \orange{ - } \red{ 3x } \green{ - } \blue{ 4 } \pink{ = } \purple{ 0 }$

$\:$

Delta

$\sf \red{ \Delta } \green{ =} \blue{ (-3) } \pink{ ^{2}} \purple{ -} \orange{ 4 } \red{ (1) } \green{ (-4) } \\ \sf \blue{ \Delta } \pink{ = } \purple{ 9 } \orange{ - } \red{ 4 } \green{ (1) } \blue{ (-4) } \\ \sf \red{\Delta }\green{=} \pink{ 9 } \purple{ -} \orange{ 4 } \red{ (-4) } \\ \sf \orange{\Delta}\pink{=} \green{ 9 } \blue{ - } \pink{ (-16) }\\ \sf \blue{\Delta}\green{=} \purple{ 9 } \orange{+ } \red{16 } \\ \sf \green{ \Delta } \blue{ = } \pink{25 } \purple{ } \orange{ } \red{ } \green{ } \blue{ } \pink{ } \purple{ } \orange{ }$

$\:$

Bhaskara

$\sf \red{ \frac{-(-3)\pm \sqrt{25}}{2(6)} } \\ \sf \green{ \frac{3\pm 5}{12} }\\ \\ \sf \blue{ \frac{3-5}{12}} \pink{= } \purple{ \frac{-2}{12} } \orange{= } \Huge\boxed{\red{ \sf \large -1} } \\ \sf \green{ \frac{3+5}{2} } \blue{ = } \pink{ \frac{8}{2} } \purple{ = } \Huge\boxed{ \orange{ \large \sf 4}} \red{ } \green{ } \blue{ } \pink{ } \purple{ } \orange{ } \red{ } \green{ } \blue{ } \pink{ } \purple{ } \orange{ } \red{ } \green{ } \blue{ } \pink{ } \purple{ } \orange{ } \red{ } \green{ } \blue{ } \pink{ } \purple{ } \orange{ }$

$\Large{\sf{\Huge ————————————————}}$

$\Large{\sf{\Huge ————————————————}}$

$\sf \red{6x^{2} } \green{ + } \blue{ 3x } \pink{ =} \purple{ 1 } \orange{+ } \red{ 2x } \\ \sf \green{ 6^{2} } \blue{ + } \pink{ 3x } \purple{ - } \orange{ 1 } \red{ - } \green{ 2x } = 0 \\ \sf \blue{ 6x^{2} } \pink{ + } \purple{ x} \orange{ - } \red{ 1} \green{ = } \blue{ 0 }$

$\:$

Delta

$\sf \Delta \red{ = } \green{ 1^{2} } \blue{ - } \pink{ 4} \purple{(6) } \orange{(-1) }\\ \sf \Delta \red{ = } \green{1 } \blue{ - } \pink{ 4 } \purple{ (6) } \orange{(1) }\\ \sf \Delta \red{ = } \green{ 1 } \blue{ -} \pink{ 4 } \purple{ (-6) }\\ \sf \Delta \orange{ = } \red{1 } \green{ -} \blue{ (-24)} \\ \sf \Delta \pink{ =} \purple{ 1 } \orange{+ } \red{ 24 } \\ \sf \Delta \green{ =} \blue{ 25 }$

$\:$

Bhaskara

$\sf \red{ \frac{-(1) \pm \sqrt{25}}{2(6)} } \\ \sf \green{ \frac{-1 \pm 5}{12} }\\ \\ \sf \blue{ \frac{-1-5}{12} } \pink{= } \purple{ \frac{-6}{12} } \orange{ =}\Huge\boxed{ \red{ \large \sf -\frac{1}{2}} } \\ \sf \green{\frac{-1+5}{12} } \blue{= } \pink{ \frac{4}{12} } \purple{ = } \Huge\boxed{ \orange{ \large \sf \frac{1}{3}} }$

Answer 2

Resposta

Resolução abaixo

Pergunta x²-2x=x+4

X²-2x=x+4

X²-2x-x-4=0

X²-3x-4=0 => temos que resolver essa equação do segundo grau

X²-3x-4=0

A=1

B=-3

C= -4

Delta

∆=b²-4.a.c

Substituindo

∆=(-3)²-4.1.(-4)

∆=9+16

∆=25

Fórmula de Bhaskara

X=-b±√∆/2.a

X=3±√25/2.1

X=3±5/2

X¹=3+5/2=> x=8/2=> x=4

X²=3-5/2=>x=-2/2=>x=-1

S(-1,4)

Bons estudos e um grande abraço

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