Question

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

Answer 1

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

Explanation:

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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