Find the zeroes of \(v^2+4\sqrt{3}v-15\) and verify relations.
\(v=\sqrt{3},\; v=-5\sqrt{3}\)
Step 1: Quadratic formula.
\(a=1, b=4\sqrt{3}, c=-15\)
Discriminant = \((4\sqrt{3})^2 - 4(1)(-15)\)
= 48 + 60 = 108
\(\sqrt{108} = 6\sqrt{3}\)
Step 2: Roots.
\(v = \dfrac{-4\sqrt{3} + 6\sqrt{3}}{2} = \sqrt{3}\)
\(v = \dfrac{-4\sqrt{3} - 6\sqrt{3}}{2} = -5\sqrt{3}\)
Step 3: Verify.
Sum = \(\sqrt{3} - 5\sqrt{3} = -4\sqrt{3}\)
= -b/a = -4\sqrt{3}
Product = \(\sqrt{3} × -5\sqrt{3} = -15\)
= c/a = -15