Prove that if the discriminant of a quadratic equation is less than zero, and the coefficient of \(x^2\) is positive then \(P(x)>0 \) for all rea values of \(x\).

In short, for polynomial \(P(x) = ax^2+bx+c\), if \(D<0\) and \(a>0\), then \(P(x)>0\) for all real values of \(x\).

Also, if \(D<0\) and \(a<0\), then, \(P(x) < 0 \).

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TopNewest\( P(x) = ax^{2} + bx + c \)

\( P(x) = a(x^{2} + \dfrac{b}{a}x + \dfrac{c}{a}) \)

\( P(x) = a\left( \left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a^{2}}\right) \)

\( P(x) = a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a} \)

Now for \( a > 0 \)

\( a\left(x+\dfrac{b}{2a}\right)^{2} \ge 0 \)

\( a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a} \ge \dfrac{4ac-b^{2}}{4a} \)

\( P(x) \ge \dfrac{4ac-b^{2}}{4a} \)

\( D < 0 \rightarrow 4ac - b^{2} > 0 \)

\( P(x) \ge \dfrac{4ac-b^{2}}{4a} > 0 \) for all x. Similar proof for \( a < 0 \) – Vighnesh Shenoy · 10 months, 2 weeks ago

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– Naitik Sanghavi · 10 months, 2 weeks ago

Perfect!Log in to reply