How do you find all the real and complex roots of #x^2 + 6x + 13 = 0#?

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Answer 1 · 2016-07-09T10:03:59

The Zeroes of given quadr. eqn. are #-3+-2i.#

There are two Methods to solve this problem :-

Method I : Method of Complete Square :

We rewrite the given eqn. as #: x^2+6x=-13.#

To make the LHS a complete square, we add

last term #= (midl. term)^2/(4*first term)=(6x)^2/(4*x^2)=9# on both sides.

#:. x^2+6x+9=-13+9#

#:. (x+3)^2=-4=(2i)^2#

#:. x+3=+-2i#

#:. x=-3+-2i#

The Zeroes of given quadr. eqn. are #-3+-2i.#

Method I : based on Formula :

The Zeroes, #alpha, beta# of a quadr. eqn. #: ax^2+bx+c=0# are #(-b+-sqrtDelta)/(2a), where, Delta= b^2-4ac.#

In our case, #a=1, b=6, c=13#, so that, #Delta =6^2-4*1*13=36-52=-16 <0..#

Hence, the roots are complex.

#alpha=(-6+sqrt(-16))/(2*1) =(-6+4i)/2=-3+2i.#

#beta=(-6-4i)/2=-3-2i.#

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