How do you solve #q(x) = -(x+2)^2+3# using the quadratic formula?

Question

Savannah Anderson · Accepted Answer

The quadratic formula is a fine way to solve this problem, but if we want to find the roots of #q#, when #q(x)=0#, this problem is already set up very well for a quick and easy algebraic manipulation to find the roots. This is essentially already in the "completed square" form if you are familiar with using completing the square to find the roots of a quadratic equation. #0=-(x+2)^2+3# Add #(x+2)^2# to both sides. #(x+2)^2=3# Take the square root of both sides, not forgetting the #pm# sign. #x+2=pmsqrt3# Subtract #2# from both sides. #x=-2pmsqrt3#

Nathan Axel · Answer

#x = -2+-sqrt3#

Extend the polynomial first.

#-(x+2)^2+3 = 0#

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

#-x^2 -4x -1 = 0#

Multiplying by -1 can simplify things.

#x^2 + 4x + 1 = 0#

Thus, it is evident that

#a=1# #b=4# #c=1#

Put these numbers into the quadratic formula last:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-4+-sqrt(4^2-4(1)(1)))/(2(1))#

#x = -2 +- sqrt12/2#

#x = -2 +- (2sqrt3)/2#

#x = -2 +- sqrt3#

Last Response

Eli Assouline · Answer

undefined To solve $ q(x) = -(x+2)^2+3 $ using the quadratic formula:

1. First, rewrite the equation in the form $ ax^2 + bx + c = 0 $:
   $ -(x+2)^2 + 3 = 0 $
   $ -x^2 - 4x - 1 = 0 $

2. Identify the values of $ a $, $ b $, and $ c $:
   $ a = -1 $, $ b = -4 $, $ c = -1 $

3. Substitute these values into the quadratic formula:
   $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

4. Plug in the values:
   $ x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(-1)(-1)}}}}{{2(-1)}} $
   $ x = \frac{{4 \pm \sqrt{{16 - 4}}}}{{-2}} $
   $ x = \frac{{4 \pm \sqrt{{12}}}}{{-2}} $

5. Simplify under the square root:
   $ x = \frac{{4 \pm 2\sqrt{{3}}}}{{-2}} $

6. Simplify further:
   $ x = -2 \pm \sqrt{3} $

Therefore, the solutions to the equation $ q(x) = -(x+2)^2+3 $ using the quadratic formula are $ x = -2 + \sqrt{3} $ and $ x = -2 - \sqrt{3} $.