Question

Solve this:
Q). If the equation $a{x}^2+bx+c$ 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,

$$\begin{gathered} Letf\left(x\right)=a{x}^2+bx+c \\ \sin cetheequationa{x}^2+bx+c=0doesnothavetwodistinctrealroots \\ Henceithastwoequalroots \\ Therefore{b}^2-4ac=0 \\ \Rightarrow b^2=4ac--\left(1\right) \\ alsogiventhata+c>b \\ squaringbothsidesweget \\ {\left(a+c\right)}^2>b^2 \\ putvalueof{b}^2=4acfromequation1 \\ \Rightarrow a^2+c^2+2ac>4ac \\ \Rightarrow a^2+c^2-2ac>0 \\ \Rightarrow {\left(a-c\right)}^2>0 \\ \Rightarrow a-c>0 \\ \Rightarrow a>c \\ \sin cea>canda+c>b \\ Henceb>a>c \\ Nowletx=-1 \\ thenf\left(-1\right)=a\times {\left(-1\right)}^2-b+c \\ =a-b+c \\ buta+c>b \\ Hencea+c-b>0 \\ Hencef\left(-1\right)>0 \\ similarlyforf(-2) \\ a{\left(-2\right)}^2-2b+c \\ \Rightarrow 4a-2b+c \\ \Rightarrow 2a+2c-2b+2a-c \\ \Rightarrow 2\left(a+c-b\right)+2a-c \\ \sin cea+c-b>0anda-c>0 \\ Hence2\left(a+c-b\right)+2a-c>0 \\ Thereforef\left(-2\right)>0 \\ Andf\left(x\right)isalwayspositiveforallpositivevaluesofx \\ Hencef\left(x\right)>0forx\in R \end{gathered}$$

Regards,