Question

Let $a,b,c\in R_0$ and 1 be a root of the equation $a{x}^2+bx+c=0$, then the equation $4a{x}^2+3bx+2c=0$ 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 $4a{x}^2+3bx+2c=0$
$D\equiv 9b^2-32ac=8(a+c{)}^2-32ac+(a+c{)}^2$
$D\equiv 9a^2+9c^2-14ac=2a^2+7a^2+2c^2+7c^2-14ac=2\left(a^2+c^2\right)+7(a-c{)}^2>0$
$\Rightarrow$ 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 $4a{x}^2+3bx+2c=0$
$D\equiv 9b^2-32ac=8(a+c{)}^2-32ac+(a+c{)}^2=9(a+c{)}^2-32ac$
$D\equiv 9a^2+9c^2-14ac(\text{ quadratic expression in ’a’) }$
$\therefore D_1=(14c{)}^2-81\left(4c^2\right)=-128c^2\Rightarrow D_1<0\Rightarrow D>0$
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 $a{x}^2+bx+c=0$ are real.
$\Rightarrow b^2-4ac\ge 0$
Now, given equation is $4a{x}^2+3bx+2c=0$
$D\equiv 9b^2-32ac=b^2+8\left(b^2-4ac\right)>0\left(\because b^2-4ac\text{ is positive }\right)$
Note that if $b^2-4ac=0$ then roots are coincident i.e. product of the roots $=1$,
$\text{ hence }c=a\text{ and }a+b+c=0\Rightarrow b=-2a$
$\therefore a=0$ (which is not possible)
Method-4: $D\equiv 9a^2+9c^2-14ac$ Now, use A.M. $>$ GM in $9a^2$ and $9c^2$ and interprect.