Question

ayuda por favor, lo pueden contestar con procedimiento, les agradecería mucho ​

Answer 1

Respuesta:

Explicación paso a paso:

1.- [tex]x^{2} -49=0[/tex]

[tex]x^{2} -7^{2} =0[/tex]

[tex](x-7)(x+7)=0\\\\\left \{ {{x-7=0} \atop {x+7=0}} \right.[/tex]

2.- [tex]3x^{2} -39=0[/tex]

[tex]3x^{2} =39[/tex]

[tex]x^{2} =\frac{39}{3} \\\\x^{2} =13\\\\\left \{ {{x1=+\sqrt{13} } \atop {x2=-\sqrt{13} }} \right.[/tex]

3.- [tex]x^{2} +25=0\\[/tex]

[tex]x^{2} +5^{2} =0\\(x+5)(x-5)=0\\\\\left \{ {{x+5=0} \atop {x-5=0}} \right.[/tex]

4.- [tex]-x^{2} =64[/tex]

[tex]x^{2} +6^{2} =0\\(x+6)(x-6)=0\\\\\left \{ {{x+6=0} \atop {x-6=0}} \right.[/tex]

5.- [tex]x^{2} +8x=0\\[/tex]

[tex]x(x+8)=0\\\\\left \{ {{x=0} \atop {x+8=0}} \right.[/tex]

6.- [tex]x^{2} =3x[/tex]

[tex]x^{2} -3x=0\\x(x-3)=0\\\\\left \{ {{x=0} \atop {x-3=0}} \right.[/tex]

7.- [tex]4x^{2} =-32x\\[/tex]

[tex]4x^{2} +32x=0\\4x(x+8)=0\\\\\left \{ {{4x=0} \atop {x+8=0}} \right.[/tex]

8.- [tex]-5x^{2} +12x=0[/tex]

[tex]5x^{2} -12x=0\\x(5x-12)=0\\\\\left \{ {{x=0} \atop {5x+12=0}} \right.[/tex]

9.- [tex]\frac{x^{2} }{6} =-x[/tex]

[tex]x^{2} =-x(6)\\x^{2} +6x=0\\x(x+6)=0\\\\\left \{ {{x=0} \atop {x+6=0}} \right.[/tex]

10.- [tex]9x^{2} =0\\[/tex]

[tex]x^{2} =\frac{0}{9}[/tex]

[tex]x^{2} =0\\x=\sqrt{0} \\x=0[/tex]

11.- [tex]4x^{2} +2=0\\[/tex]

[tex]4x^{2} =0-2\\x^{2} =\frac{-2}{4} \\\\\left \{ {{x =\sqrt{\frac{-2}{4} } } \atop {x=-\sqrt{\frac{-2}{4} } }} \right.[/tex]