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Scilla [17]
3 years ago
13

Solve.

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
7 0
<span>Solve.

x² + 4x + 4 = 18


​ A x=−4±3√ ​2

​ B x=2±3√ ​2

​ C x=4±9√ ​2

D x=−2±3√2



</span>x^2 + 4x + 4 = 18 \\  \\ x^2+4x+4-18= 0 \\  \\ x^2+4x-14=0 \\  \\ x_1_y_2=  \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{16-(-56)} }{2}  \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{72} }{2}  \\  \\  \\  x_1=  \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2=  \dfrac{-4+ \sqrt{72} }{2}\\  \\  \\  x_1=  \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2=  \dfrac{-4- \sqrt{2^3*3^2} }{2}
<span>
</span>x_1=  \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 2*3\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 6\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2=   \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\  \\  \\  x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad  x_2=  -2- 3\sqrt{2}  \\  \\  \\ \boxed{x=   -2\pm 3\sqrt{2} }  \to D)<span>
</span>
tatyana61 [14]3 years ago
7 0

This was the correct answer, I even provided proof from my own test. Hope this helps!

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