Question 111575

Question 111575

Free

Find the roots of the equation \(a{{x}^{2}}+bx+c=0\) in terms of \(a\), \(b\) and \(c\).

Assume \(a\ne0\).

Default solution

The equation is already in the form \(ax^2+bx+c=0\), with \(a\ne0\). The quadratic formula gives

\(x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\quad\text{or}\quad x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\)

For real coefficients, these roots are real when \(b^2-4ac\ge0\). If \(b^2-4ac=0\), the two values coincide at \(x=-\dfrac{b}{2a}\). If \(b^2-4ac<0\), there are no real roots.

Answer:\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), with real roots when \(b^2-4ac\ge0\); if \(b^2-4ac=0\), the repeated root is \(-\dfrac{b}{2a}\).

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