Respuesta :
[tex]( {x}^{2} + 6x + 8)( {x}^{2} + 6x + 13) =0 \\ [/tex]
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[tex] {x}^{2} + 6x + 8 = 0 [/tex]
[tex](x + 2)(x + 4) = 0[/tex]
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[tex]x + 2 = 0[/tex]
[tex]x = - 2[/tex]
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[tex]x + 4 = 0[/tex]
[tex]x = - 4[/tex]
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[tex] {x}^{2} + 6x + 13 = 0 [/tex]
[tex]∆ = {b}^{2} - 4ac [/tex]
[tex]a = coefficient \: \: of \: \: {x}^{2} = 1 [/tex]
[tex]b = coefficient \: \: of \: \: x = 6[/tex]
[tex]c = the \: \: alone \: \: number \: = 13 \\ [/tex]
Thus ;
[tex]∆ = ({6})^{2} - 4 \times (2) \times (13) [/tex]
[tex]∆ = 36 - 104[/tex]
[tex]∆ = - 68[/tex]
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Point :
Remember from now on ,
In quadratic functions ;
if :
[tex]∆ > 0[/tex]
The function has two roots
if :
[tex]∆ = 0[/tex]
The function has just one root
if :
[tex]∆ < 0[/tex]
The function doesn't have any root.
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Thus , ( x² + 6x + 13 ) doesn't have any root.
So ; ( x = - 2 ) & ( x = - 4 ) are the only roots.
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And we're done....♥️♥️♥️♥️♥️
Answer:
c)-2,-4,-3+2i,-3-2i
Step-by-step explanation:
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