Question #429db

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Answer 1 · 2017-03-20T11:37:24

# k=+-9.#

Let #alpha & beta# be the zeroes of the Quadr. #ax^2+bx+c=0.#

By the quadr. form., we have, then,

#alpha, beta=(-b+-sqrt(b^2-4ac))/(2a).#

# rArr |alpha-beta|=1/(2a){(-b+sqrt(b^2-4ac))-(-b-sqrt(b^2-4ac))}#

#=1/(2a)(2sqrt(b^2-4ac))#

#:. |alpha-beta|=sqrt(b^2-4ac)/a...............(ast).#

In our Problem, we have,

#|alpha-beta|=sqrt69, a=1, b=k, c=3.#

By #(ast)," then, "sqrt69=sqrt{k^2-4(1)(3)}=sqrt(k^2-12).#

# rArr k^2-12=69, or, k^2=81.#

#:. k=+-9#.

Enjoy Maths.!