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.!