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

Ax= -8+5 square root 2
Bx=-6+5 square root 2
Cx= -4+5 square root 2
Dx= -2+5 square root 2

Verified Answer :

Petroniusz

(x + 2)² + 12(x + 2) – 14 = 0

(x +2) ² we use the formula (a+b)² = a² + 2ab + b²

(x² + 22x + 4 ) + (12x + 12 2 ) - 14 = 0
(x² + 4x + 4) + (12x + 24) - 14 = 0
x² + 4x + 4 + 12x + 24 - 14 = 0

we sort expressions
x² + 4x + 12x + 4 + 24 - 14 = 0

reduce
x² + 16x + 14 = 0 

a = 1    b = 16     c=14

Δ = b² - 4ac = 16² - 4114 = 256 - 56 = 200
√Δ = √(21010) = 10√2

\(x_{1} = \frac{-b - \sqrt{delty} }{2a} = \frac{-16 - 10 \sqrt{2} }{2*1} = \frac{-8 -5 \sqrt{2} }{1} = -8 -5 \sqrt{2}\)

\(x_{2} = \frac{-b + \sqrt{delty} }{2a} = \frac{-16+ 10 \sqrt{5} }{2*1} = \frac{-8 + 5 \sqrt{2} }{1} = -8 + 5 \sqrt{2}\)

Answer A

The solution to the equation is x = -8 ± 5√2, the correct option is A.

What is a Quadratic Equation?

A quadratic equation is what can be written in the form of ax²+bx+c=0

The quadratic equation is

(x + 2)² + 12(x + 2) – 14 = 0?

Let y = x+2

then the equation will be of the form

y² +12y -14 = 0

This in the form of ax² +bx+c =0

The formula for solving is,

\(\rm y =\dfrac{-b \pm\sqrt{b^2 -4ac}}{2a}\)

\(\rm y= \dfrac{ -12 \pm \sqrt {(-12)^2 -4 *1*(-14)}}{ 2* 1}\)

y = -6 ± 5√2

x +2 = -6 ± 5√2

x = -8 ± 5√2

To know more about Quadratic Equation

https://brainly.com/question/2263981

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