Respuesta :

frika

Answer:

[tex]\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}[/tex]

Step-by-step explanation:

For the equation [tex](x+4)^2-3(x+4)-3=0[/tex] use the substitution [tex]u=x+4.[/tex] Then the equation will take look

[tex]u^2-3u-3=0.[/tex]

Solve this quadratic equation:

[tex]D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.[/tex]

Thus,

[tex]x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.[/tex]

zainebamir540

Answer:

x = (-5+√21) /2    or x = (-5-√21) /2

Step-by-step explanation:

We have given  the equation :

(x + 4)²– 3(x + 4) – 3=0

Put u = (x + 4) in the given equation we get,

u² -3u -3 = 0

Using quadratic equation we get,

u =  ( -(-3)±√(-3)²-4(1)(-3) ) /2(1)

u = (3±√21) /2

Then replace u by (x+4) we get,

(x+4) = (3±√21) /2

x+4 = (3±√21) /2

x = (3(3±√21) /2 +(-4)

x = (-5±√21) /2

x = (-5+√21) /2    or x = (-5-√21) /2 is the solution.

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