Respuesta :
Let u = x+2. That makes our equation, in terms of u: [tex]u^2+12u-14=0[/tex]. Fitting this into the quadratic formula looks like this: [tex]u= \frac{-12+/- \sqrt{12^2-4(1)(-14)} }{2} [/tex]. Simplifying a bit gives us [tex]u= \frac{-12+/- \sqrt{200} }{2} [/tex]. Simplifying that in terms of the radical gives us [tex]u= \frac{-12+/-10 \sqrt{2} }{2} [/tex]. Reducing that numerator by the 2 in the denominator gives us [tex]u=-6+5 \sqrt{2},-6-5 \sqrt{2} [/tex]. If u = x + 2, then we make that substitution now to solve for x: [tex]x+2=-6+5 \sqrt{2} [/tex] and [tex]x=-10+5 \sqrt{2} [/tex]; [tex]x+2=-6-5 \sqrt{2} [/tex] and [tex]x=-10-5 \sqrt{2} [/tex]
Answer:
[tex]\text{The solutions are}x=-8+5\sqrt2,-8-5\sqrt2[/tex]
Step-by-step explanation:
Given the equation
[tex](x+2)^2+12(x+2)-14=0[/tex]
we have to find the solution of above equation by using quadratic formula.
[tex](x+2)^2+12(x+2)-14=0[/tex]
Substituting x+2=u, we get
Equation:[tex]u^2+12u-14=0[/tex]
[tex]\text{Comparing above with standard equation }ax^2+bx+c=0\text{, we get}[/tex]
a=1, b=12, c=-14
By quadratic formula
[tex]a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]a=\frac{-12\pm \sqrt{(12)^2-4(1)(-14)}}{2(1)}[/tex]
[tex]a=\frac{-12\pm \sqrt{200}}{2}[/tex]
[tex]a=\frac{-12\pm 10\sqrt{2}}{2}[/tex]
[tex]a=-6+5\sqrt{2},-6-5\sqrt{2}[/tex]
[tex]\text{The solutions for u are }-6+5\sqrt{2},-6-5\sqrt{2}[/tex]
⇒ [tex]x+2=-6+5\sqrt{2}[/tex] and [tex]x+2=-6-5\sqrt2[/tex]
[tex]x=-8+5\sqrt2,-8-5\sqrt2[/tex]
Option A is correct
