Everything is in 'k'

Geometry Level 4

The smallest value of k, for which both the roots of the equation x 2 − 8 k x + 16 ( k 2 − k + 1 ) = 0 { x }_{ }^{ 2 }-8kx+16(k{ }_{ }^{ 2 }-k+1)=0 are real, distinct and have values at least 4, is


The answer is 2.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Using quadratic formula, 8 K ± 64 K 2 − 64 K 2 + 64 K − 64 2 = 4 K ± 4 K − 1 . \dfrac {8K\pm\ \sqrt{64K^2-64K^2+64K-64} } 2=4K\ \pm\ 4\sqrt{K -1}.
If K=1, roots are not distinct. If K=2, all conditions are fulfilled.
So K= 2 \ \ \color{#D61F06}{2}

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