How do you find the roots, real and imaginary, of #y=(x +5 )^2+6x+3# using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

#x1 = -14#
#x2 = -2# #y = (x + 5)^2 +6x +3# Expand square term: #y = x^2 + 10x + 25 + 6x + 3# Therefore #y = x^2 +16x + 28# This quadratic factorizes so there is no need of the Quadratic Formula: #y = (x + 14)(x + 2)# which has zeros at # x = -14 and -2# If you would like to use the Quadratic Formula: #x = (-b +- sqrt(b^2 - 4ac))/(2a)# In this case #a = 1, b = 16, c = 28# Hence; #x = (-16 +- sqrt(256 - 112))/2# #x = (-16+- sqrt(144))/2# #x= (-16 +- 12)/2# #x = -14 or -2# Regarding your question on real or complex roots, the roots of a quadratic function with real coefficients will either both be real or both be complex. In this case the roots are real.

Abigail Allen · Answer

undefined To find the roots of $ y = (x + 5)^2 + 6x + 3 $ using the quadratic formula:

1. Identify the coefficients: $ a = 1 $, $ b = 6 $, and $ c = 3 $.
2. Apply the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $.
3. Substitute the values of $ a $, $ b $, and $ c $ into the formula.
4. Solve for $ x $.

$ x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 1 \cdot 3}}}}{{2 \cdot 1}} $.

$ x = \frac{{-6 \pm \sqrt{{36 - 12}}}}{2} $.

$ x = \frac{{-6 \pm \sqrt{24}}}{2} $.

$ x = \frac{{-6 \pm 2\sqrt{6}}}{2} $.

$ x = -3 \pm \sqrt{6} $.

So, the roots of the equation are $ x = -3 + \sqrt{6} $ and $ x = -3 - \sqrt{6} $.