Respuesta :
Answer:
[tex]x=-4 \pm 2\sqrt{3}[/tex]
Step-by-step explanation:
The zeros of f is when f=0.
So we need to solve:
[tex]x^2+8x+4=0[/tex]
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I'm going to choose completing the square.
Subtract 4 on both sides:
[tex]x^2+8x=-4[/tex]
Add (8/2)^2 on both sides:
[tex]x^2+8x+(\frac{8}{2})^2=-4+(\frac{8}{2})^2[/tex]
Write left hand side as a square:
[tex](x+\frac{8}{2})^2=-4+4^2[/tex]
[tex](x+4)^2=-4+16[/tex]
[tex](x+4)^2=12[/tex]
Take the square root of both sides:
[tex](x+4)=\pm \sqrt{12}[/tex]
[tex]x+4=\pm \sqrt{12}[/tex]
Simplify right hand side:
[tex]x+4=\pm \sqrt{4}\sqrt{3}[/tex]
[tex]x+4=\pm 2\sqrt{3}[/tex]
Subtract 4 on both sides:
[tex]x=-4 \pm 2\sqrt{3}[/tex]
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You could also go with quadratic formula:
[tex]a=1[/tex]
[tex]b=8[/tex]
[tex]c=4[/tex]
We need to use this formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
I'm going to evaluate [tex]b^2-4ac[/tex] first.
[tex]b^2-4ac=(8)^2-4(1)(4)=64-16=48[/tex]
So now we have this so for with the formula:
[tex]x=\frac{-8 \pm \sqrt{48}}{2(1)}[/tex]
Simplifying the denominator gives:
[tex]x=\frac{-8 \pm \sqrt{48}}{2}[/tex]
Now is there a perfect square in 48? Yes, 16 is a perfect square factor in 48.
So we can write:
[tex]x=\frac{-8 \pm \sqrt{16}\sqrt{3}}{2}[/tex]
[tex]x=\frac{-8 \pm 4\sqrt{3}}{2}[/tex]
[tex]x=\frac{-8}{2}\pm \frac{4\sqrt{3}}{2}[/tex]
Reduce fractions:
[tex]x=-4 \pm 2\sqrt{3}[/tex]
