Find the zeroes of \(2x^2+\dfrac{7}{2}x+\dfrac{3}{4}\) and verify relations.
\(x=-\dfrac{1}{4},\; x=-\dfrac{3}{2}\)
Step 1: Remove fractions.
Multiply whole equation by 4:
\(8x^2 + 14x + 3\)
Step 2: Factorise.
\(8x^2 + 14x + 3 = (4x + 1)(2x + 3)\)
Step 3: Zeroes.
\(4x + 1 = 0 \Rightarrow x = -\dfrac{1}{4}\)
\(2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}\)
Step 4: Verify.
Sum = \(-\dfrac{1}{4} - \dfrac{3}{2} = -\dfrac{7}{4}\)
= \(-b/a = - (7/2)/2 = -7/4\)
Product = \((-\dfrac{1}{4})(-\dfrac{3}{2}) = 3/8\)
= \(c/a = (3/4)/2 = 3/8\)