Find the zeroes of \(7y^2-\dfrac{11}{3}y-23\) and verify relations.
\(y=\dfrac{11+\sqrt{5917}}{42},\; y=\dfrac{11-\sqrt{5917}}{42}\)
Step 1: Identify coefficients.
\(a = 7,\; b = -\dfrac{11}{3},\; c = -23\)
Step 2: Discriminant.
\(b^2 - 4ac = \left(-\dfrac{11}{3}\right)^2 - 4(7)(-23)\)
= \(\dfrac{121}{9} + 644\)
= \(\dfrac{5917}{9}\)
So, \(\sqrt{\Delta} = \dfrac{\sqrt{5917}}{3}\)
Step 3: Roots.
\(y = \dfrac{ -b \pm \sqrt{\Delta} }{ 2a }\)
= \(\dfrac{ \dfrac{11}{3} \pm \dfrac{\sqrt{5917}}{3} }{ 14 }\)
= \(\dfrac{ 11 \pm \sqrt{5917} }{ 42 }\)
Step 4: Verify.
Sum = \(\dfrac{11 + \sqrt{5917}}{42} + \dfrac{11 - \sqrt{5917}}{42}\)
= \(\dfrac{22}{42} = \dfrac{11}{21}\)
= \(-\dfrac{b}{a} = -\dfrac{-11/3}{7} = \dfrac{11}{21}\)
Product = \(\dfrac{(11 + \sqrt{5917})(11 - \sqrt{5917})}{42^2}\)
= \(\dfrac{121 - 5917}{1764}\)
= \(\dfrac{-5796}{1764}\)
= \(-\dfrac{23}{7}\)
= \(\dfrac{c}{a}\)