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

2 Answers
Jun 11, 2017

Answer:

#x = -2+-sqrt3#

Explanation:

First, expand the polynomial.

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

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

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

You can multiply by -1 to make things easier.

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

So we see that

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

Finally, plug these values into the quadratic formula:

#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#

Final Answer

Jun 12, 2017

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.

If you're familiar with completing the square to find the roots of a quadratic equation, this is essentially already in the "completed square" form.

#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#