Solving Quadratic Equations

Solve the equations:

(a) \(3x^{2}-5x+1=0\)

(b) \(x^{2}+x=-1\)

Solution (a) The discriminant: \(b^{2}-4ac=25-12=13\). We use the quadratic formula. \[ x=\dfrac{-b\pm \displaystyle\sqrt{b^{2}-4ac}}{2a}=\dfrac{5\pm \displaystyle\sqrt{13}}{6} \]

The solutions are \(x=\dfrac{5-\sqrt{13}}{6}\approx 0.232\) and \(x=\dfrac{5+ \sqrt{13}}{6}\approx 1.434\)

(b) The equation in standard form is \(x^2+x+1=0\). Its discriminant is \(b^{2}-4ac=1-4=-3\). This equation has no real solution.