Question

Let $a, b, c \in R_0$ and 1 be a root of the equation $a x^2+b x+c=0$, then the equation $4 a x^2+3 b x+2 c=0$ has

• A. imaginary roots
• B. real and equal roots
• C. real and unequal roots
• D. rational roots

Answer 1

Answer: (C)

Method-1: As $x=1$ is root, so $a+b+c=0 \Rightarrow b=-(a+c)$
Now, given equation is $4 a x^2+3 b x+2 c=0$
$D \equiv 9 b ^2-32 ac =8( a + c )^2-32 ac +( a + c )^2 $
$D \equiv 9 a ^2+9 c ^2-14 ac =2 a ^2+7 a ^2+2 c ^2+7 c ^2-14 ac =2\left( a ^2+ c ^2\right)+7( a - c )^2>0$
$\Rightarrow$ roots are real and distinct.
Method-2:As $x =1$ is root, so $a + b + c =0 \Rightarrow b =-( a + c )$
Now, given equation is $4 a x^2+3 b x+2 c=0$
$D \equiv 9 b^2-32 a c=8(a+c)^2-32 a c+(a+c)^2=9(a+c)^2-32 a c $
$D \equiv 9 a^2+9 c^2-14 a c(\text { quadratic expression in 'a') }$
$\therefore D_1=(14 c)^2-81\left(4 c^2\right)=-128 c^2 \Rightarrow D_1<0 \Rightarrow D>0$
So, the equation has real and unequal roots. Ans.
Method-3: Since $x=1$ is one of the roots and therefore roots of the equation $ax ^2+ bx + c =0$ are real.
$\Rightarrow b ^2-4 ac \geq 0$
Now, given equation is $4 a x^2+3 b x+2 c=0$
$D \equiv 9 b^2-32 a c=b^2+8\left(b^2-4 a c\right)>0 \left(\because b^2-4 a c \text { is positive }\right)$
Note that if $b^2-4 a c=0$ then roots are coincident i.e. product of the roots $=1$,
$\text { hence } c = a \text { and } a + b + c =0 \Rightarrow b =-2 a$
$\therefore a=0$ (which is not possible)
Method-4: $D \equiv 9 a^2+9 c^2-14 a c$ Now, use A.M. $>$ GM in $9 a^2$ and $9 c^2$ and interprect.