Solve.x² + 4x + 4 = 18 A x=−4±3√ 2 B x=2±3√ 2 C x=4±9√ 2D x=−2±3√2
Question
Answer:
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
[tex]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}[/tex]
[tex] 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) [/tex]
solved
general
11 months ago
4705