Question

descubra o valor de x em 4x²-5=2(2x+5)​

Answer 1

Resposta

$4x^2-5=2(2x+5)\\ \\ 4x^2-5=4x+10\\ \\ 4x^2-4x-5-10=0\\ \\ 4x^2-4x-15=0$

"método de Bhaskara"

coeficientes

a = 4

b= -4

c = -15

Δ = b² -4ac

Δ = ( -4)² -4(4)(-15)

Δ = 16+240

Δ = 256

$x=\dfrac{-b\pm\sqrt{\Delta} }{2a}\\ \\ \\ x=\dfrac{-(-4)\pm\sqrt{256}}{2(4)}=\dfrac{4\pm16}{8}\\ \\ \\ x'=\dfrac{4+16}{8}=\dfrac{20}{8}=\boxed{\dfrac{5}{4}}\\ \\ \\ x"=\dfrac{4-16}{8}=-\dfrac{12}{8}=\boxed{-\dfrac{3}{2}}$