How do you solve #x^2 + 10x + 25 = 0# using the quadratic formula?

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Answer 1 · 2017-05-22T02:13:23

#x = -5#

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

Where #ax^2+bx+c=0#.

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In this case, #a = 1#, #b=10#, and #c=25#. Now all we have to do is plug these values into the quadratic formula and simplify it.

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

#x = (-10+-sqrt(10^2-4(1)(25)))/(2(1))#

#x = (-10+-sqrt(100-100))/2#

#x = (-10+-0)/2#

Since adding or subtracting #0# wouldn't change the answer at all, we actually only have a single solution instead of 2 solutions.

#x = (-10)/2 = -5#

Final Answer

Answer 2 · 2017-05-22T04:21:53

Double root at x = - 5

#y = x^2 + 10x + 25 = 0#
#D = b^2 - 4ac = 100 - 100 = 0#
Since D = 0, there is a double root at:
#x = -b/(2a) = -10/2 = - 5#