Which equation has the sum of its roots equal to \(3\)?
\(2x^2-3x+6=0\)
\(-x^2+3x-3=0\)
\(\sqrt{2}\,x^2-\dfrac{3}{\sqrt{2}}x+1=0\)
\(3x^2-3x+3=0\)
Step 1: The general form of a quadratic equation is \(ax^2 + bx + c = 0\).
Step 2: The formula for the sum of the roots is:
\[ \text{Sum of roots} = -\dfrac{b}{a} \]
Here, \(a\) is the coefficient of \(x^2\), and \(b\) is the coefficient of \(x\).
Step 3: Check each option one by one.
(A) \(2x^2 - 3x + 6 = 0\): Here, \(a = 2\), \(b = -3\). So, \(-b/a = -(-3)/2 = 3/2\). Not equal to 3.
(B) \(-x^2 + 3x - 3 = 0\): Here, \(a = -1\), \(b = 3\). So, \(-b/a = -(3)/(-1) = 3\). Yes, this matches the required sum.
(C) \(\sqrt{2}\,x^2 - \dfrac{3}{\sqrt{2}}x + 1 = 0\): Here, \(a = \sqrt{2}\), \(b = -\dfrac{3}{\sqrt{2}}\). So, \(-b/a = -\left(-\dfrac{3}{\sqrt{2}}\right) / \sqrt{2} = \dfrac{3}{2}\). Not equal to 3.
(D) \(3x^2 - 3x + 3 = 0\): Here, \(a = 3\), \(b = -3\). So, \(-b/a = -(-3)/3 = 1\). Not equal to 3.
Step 4: Therefore, the correct option is B.