Question

What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to solve.

Answer 1

Answer:

$\text{The solutions are}x=-8+5\sqrt2,-8-5\sqrt2$

Step-by-step explanation:

Given the equation

$(x+2)^2+12(x+2)-14=0$

we have to find the solution of above equation by using quadratic formula.

$(x+2)^2+12(x+2)-14=0$

Substituting x+2=u, we get

Equation:$u^2+12u-14=0$

$\text{Comparing above with standard equation }ax^2+bx+c=0\text{, we get}$

a=1, b=12, c=-14

By quadratic formula

$a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=\frac{-12\pm \sqrt{(12)^2-4(1)(-14)}}{2(1)}$

$a=\frac{-12\pm \sqrt{200}}{2}$

$a=\frac{-12\pm 10\sqrt{2}}{2}$

$a=-6+5\sqrt{2},-6-5\sqrt{2}$

$\text{The solutions for u are }-6+5\sqrt{2},-6-5\sqrt{2}$

⇒ $x+2=-6+5\sqrt{2}$ and $x+2=-6-5\sqrt2$

$x=-8+5\sqrt2,-8-5\sqrt2$

Option A is correct

Answer 2

Let u = x+2.  That makes our equation, in terms of u:  $u^2+12u-14=0$.  Fitting this into the quadratic formula looks like this:  $u= \frac{-12+/- \sqrt{12^2-4(1)(-14)} }{2}$.  Simplifying a bit gives us  $u= \frac{-12+/- \sqrt{200} }{2}$.  Simplifying that in terms of the radical gives us  $u= \frac{-12+/-10 \sqrt{2} }{2}$.  Reducing that numerator by the 2 in the denominator gives us  $u=-6+5 \sqrt{2},-6-5 \sqrt{2}$.  If u = x + 2, then we make that substitution now to solve for x:  $x+2=-6+5 \sqrt{2}$  and  $x=-10+5 \sqrt{2}$;  $x+2=-6-5 \sqrt{2}$  and  $x=-10-5 \sqrt{2}$