How do you solve #x^2 + 4x + 4 = 7# and simplify the answer in simplest radical form?

1 Answer

See a solution process below:

Explanation:

First, subtract #color(red)(7)# from each side of the equation to put the equation in standard form:

#x^2 + 4x + 4 - color(red)(7) = 7 - color(red)(7)#

#x^2 + 4x - 3 = 0#

We can now use the quadratic formula to find the solutions for #x#. The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(1)# for #color(red)(a)#

#color(blue)(4)# for #color(blue)(b)#

#color(green)(-3)# for #color(green)(c)# gives:

#x = (-color(blue)(4) +- sqrt(color(blue)(4)^2 - (4 * color(red)(1) * color(green)(-3))))/(2 * color(red)(1))#

#x = (-color(blue)(4) +- sqrt(16 - (-12)))/2#

#x = (-color(blue)(4) +- sqrt(16 + 12))/2#

#x = (-color(blue)(4) +- sqrt(28))/2#

#x = (-color(blue)(4) +- sqrt(4 * 7))/2#

#x = (-color(blue)(4) +- sqrt(4)sqrt(7))/2#

#x = (-color(blue)(4) +- 2sqrt(7))/2#

#x = -color(blue)(4)/2 +- (2sqrt(7))/2#

#x = -2 +- sqrt(7)#

Or

#x = -2 + sqrt(7)# and #x = -2 - sqrt(7)#