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Let $f(x)=3ax^2−4bx+c$ $(a,b,c∈R,a
eq 0)$ where $a,\ b,\ c$ are in A.P. Then how many roots the equation $f(x)=0$$ have? Are they real?
Given: $f(x)=3ax^2−4bx+c$ $(a,b,c∈R,a\
eq 0)$ where $a,\ b,\ c$ are in A.P.
eq 0)$ where $a,\ b,\ c$ are in A.P.
To do: To find the number of roots of $f( x)=0$ and to know weather they are real or not.
Solution:
$\because a,\ b,\ c$ are in A.P.,
$\therefore 2b=a+c$
$4b^2=( a+c)^2$ [on squaring both sides]
The discriminant of the given function $f( x)=3ax^2−4bx+c$ is,
$D=16b^2−12ac$
$=4( a+c)^2−12ac$
$=4[( a^2+c^2+2ac)−3ac]$
$=4( a^2+c^2−ac)$
$=4( a^2+c^2−2ac+ac)$
$=4( ( a−c)^2+ac)$
Case 1: If $a$ and $c$ are of opposite signs, then, $D=(+)ve$.
Case 2: If $a$ and $c$ are of same signs, then, $D=(+)ve$.
This shows that $f(x)=0$ has two unequal real roots.
 
Updated on: 10-Oct-2022
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