Question

Q.1. If the equation $a{x}^2+bx+c=0$ does not have 2 distinct real roots and a + c > b, then prove that $f\left(x\right)\ge 0,\forall x\in R$

Answer 1

Dear student

$$\begin{gathered} \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}, \\ \mathrm{given}\mathrm{a}+\mathrm{c}>\mathrm{b}\Rightarrow \mathrm{a}+\mathrm{c}-\mathrm{b}>0...\left(1\right) \\ \mathrm{Since}\mathrm{roots}\mathrm{are}\mathrm{non}\mathrm{real} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)<0 \\ \mathrm{Now},\mathrm{f}\left(-1\right)=\mathrm{a}{\left(-1\right)}^2+\mathrm{b}\left(-1\right)+\mathrm{c} \\ =\mathrm{a}+\mathrm{c}-\mathrm{b} \\ >0\left[\mathrm{using}\left(1\right)\right] \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)\ge 0\mathrm{for}\mathrm{all}\mathrm{x}\in \mathrm{R} \end{gathered}$$
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