Respuesta :
Answer:
x = 3 + √6 ; x = 3 - √6 ; [tex]x = \frac{2+3\sqrt{2}}{2}[/tex] ; [tex]x = \frac{2-(3)\sqrt{2}}{2}[/tex]
Step-by-step explanation:
Relation given in the question:
(x² − 6x +3)(2x² − 4x − 7) = 0
Now,
for the above relation to be true the following condition must be followed:
Either (x² − 6x +3) = 0 ............(1)
or
(2x² − 4x − 7) = 0 ..........(2)
now considering the equation (1)
(x² − 6x +3) = 0
the roots can be found out as:
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
for the equation ax² + bx + c = 0
thus,
the roots are
[tex]x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}[/tex]
or
[tex]x = \frac{6\pm\sqrt{36-12}}{2}[/tex]
or
[tex]x = \frac{6+\sqrt{24}}{2}[/tex] and, x = [tex]x = \frac{6-\sqrt{24}}{2}[/tex]
or
[tex]x = \frac{6+2\sqrt{6}}{2}[/tex] and, x = [tex]x = \frac{6-2\sqrt{6}}{2}[/tex]
or
x = 3 + √6 and x = 3 - √6
similarly for (2x² − 4x − 7) = 0.
we have
the roots are
[tex]x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}[/tex]
or
[tex]x = \frac{4\pm\sqrt{16+56}}{4}[/tex]
or
[tex]x = \frac{4+\sqrt{72}}{4}[/tex] and, x = [tex]x = \frac{4-\sqrt{72}}{4}[/tex]
or
[tex]x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}[/tex] and, x = [tex]x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}[/tex]
or
[tex]x = \frac{4+(2\times3)\sqrt{2}}{2}[/tex] and, x = [tex]x = \frac{4-(2\times3)\sqrt{2}}{4}[/tex]
or
[tex]x = \frac{2+3\sqrt{2}}{2}[/tex] and, [tex]x = \frac{2-(3)\sqrt{2}}{2}[/tex]
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ; [tex]x = \frac{2+3\sqrt{2}}{2}[/tex] ; [tex]x = \frac{2-(3)\sqrt{2}}{2}[/tex]
