Question

ayuda con la siguientes ecuaciones cuadraticas xfavor

Answer 1

Respuesta:

a)
$x_{1}=-4\\ \\x_{2}=-2$
b)
$x_{1} =-6\\\\x_{2} =8$
c)
$x_{1}=-2\\ \\x_{2} =0$
d)
$x_{1} =-3\\\\x_{2} =0$
e)
$x_{1}=-3\\ \\x_{2}=3$
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Explicación paso a paso:

Utilizamos la fórmula general e las a), b), c) y d) que es
$ax^{2} +bx+c\\\\x_{1} =\frac{-b-\sqrt{b^{2}-4ac } }{2a} \\\\x_{2} =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$x^{2} +6x+8\\\\x_{1}=\frac{-6-\sqrt{6^{2}-4(1)(8) } }{2(1)}\\ \\x_{1}=\frac{-6-\sqrt{36-32 } }{2}\\ \\x_{1}=\frac{-6-\sqrt{4} }{2}\\\\x_{1}=\frac{-6-2 }{2}\\\\x_{1}=\frac{-8}{2}\\ \\x_{1}=-4\\\\x_{2} =\frac{-6+\sqrt{6^{2}-4(1)(8) } }{2(1)}\\\\x_{2} =\frac{-6+\sqrt{36-32 } }{2}\\\\x_{2} =\frac{-6+\sqrt{4 } }{2}\\\\x_{2} =\frac{-6+2 }{2}\\\\x_{2} =\frac{-4 }{2}\\\\x_{2} =-2$
$x^{2} -2x-48\\\\\\x_{1} =\frac{-(-2)-\sqrt{(-2)^{2}-4(1)(-48) } }{2(1)} \\\\x_{1} =\frac{2-\sqrt{(4+192) } }{2} \\\\x_{1} =\frac{2-\sqrt{196 } }{2} \\\\x_{1} =\frac{2-14 }{2} \\\\x_{1} =\frac{-12 }{2} \\\\x_{1} =-6\\\\x_{2} =\frac{-(-2)+\sqrt{(-2)^{2}-4(1)(-48) } }{2(1)} \\\\x_{2} =\frac{2+\sqrt{(4+192) } }{2} \\\\x_{2} =\frac{2+\sqrt{196 } }{2} \\\\x_{2} =\frac{2+14 }{2} \\\\x_{2} =\frac{16 }{2}\\\\x_{2} =8$
$4x^{2} +8x=0\\\\x_{1} =\frac{-8-\sqrt{8^{2}-4(4)(0) } }{2(4)} \\\\x_{1} =\frac{-8-\sqrt{64 } }{8} \\\\x_{1} =\frac{-8-8 }{8} \\\\x_{1} =\frac{-16 }{8} \\\\x_{1} =-2 \\\\x_{2} =\frac{-8+\sqrt{8^{2}-4(4)(0) } }{2(4)} \\\\x_{2} =\frac{-8+\sqrt{64 } }{8} \\\\x_{2} =\frac{-8+8 }{8} \\\\x_{2} =\frac{0 }{8} \\\\x_{2} =0$
$x^{2} +3x=0\\\\x_{1} =\frac{-3-\sqrt{3^{2}-4(1)(0) } }{2(1)} \\\\x_{1} =\frac{-3-\sqrt{9 } }{2} \\\\x_{1} =\frac{-3-3 }{2} \\\\x_{1} =\frac{-6 }{2}\\ \\x_{1} =-3\\\\x_{2} =\frac{-3+\sqrt{3^{2}-4(1)(0) } }{2(1)} \\\\x_{2} =\frac{-3+\sqrt{9 } }{2} \\\\x_{2} =\frac{-3+3 }{2} \\\\x_{2} =\frac{0 }{2}\\ \\x_{2} =0$

$x^{2} -9=0\\\\x^{2} =9\\\\x_{1} = -\sqrt{9} \\\\x_{1} = -3\\\\x_{2}=\sqrt{9} \\\\x_{2}=3$