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.

a. x= -8+5 square root 2b) x=-6+5 square root 2c) x= -4+5 square root 2d) x= -2+5 square root 2?

Answer 1

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= (-12+/- √(12^2-4(1)(-14)) )/(2)`. Simplifying a bit gives us
`u= (-12+/- √(200) )/(2)`. Simplifying that in terms of the radical gives us
`u= (-12+/-10 √(2) )/(2)`. Reducing that numerator by the 2 in the denominator gives us
`u=-6+5 √(2),-6-5 √(2)`. If u = x + 2, then we make that substitution now to solve for x:
`x+2=-6+5 √(2)` and
`x=-10+5 √(2)`;
`x+2=-6-5 √(2)` and
`x=-10-5 √(2)`

Answer 2

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

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=(-b\pm √(b^2-4ac))/(2a)`

`a=(-12\pm √((12)^2-4(1)(-14)))/(2(1))`

`a=(-12\pm √(200))/(2)`

`a=(-12\pm 10√(2))/(2)`

`a=-6+5√(2),-6-5√(2)`

`\text{The solutions for u are }-6+5√(2),-6-5√(2)`

⇒
`x+2=-6+5√(2)` and
`x+2=-6-5\sqrt2`

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

Option A is correct