Question

If $2x^2+5x+2b=0$ and $2x^3+7x^2+5x+1=0$ have atleast one common root for three values of $b$, then the sum of all three values of $b$ is

• A. $\frac{1}{2}$
• B. $0$
• C. $\frac{5}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

$2x^2+5x+2b=0\Rightarrow 2x^2+5x=-2b\cdots \left(1\right)2x^3+7x^2+5x+1=0\Rightarrow 2x^3+5x^2+2x^2+5x+1=0$
Using equation $\left(1\right)$, we get
$\Rightarrow x(-2b)-2b+1=0\Rightarrow x=\frac{1-2b}{2b}$
Putting this in equation $\left(1\right)$, we get
$2{\left(\frac{1-2b}{2b}\right)}^2+5\left(\frac{1-2b}{2b}\right)+2b=0\Rightarrow \frac{4b^2-4b+1+5b-10b^2+4b^3}{2b^2}=0\Rightarrow \frac{4b^3-6b^2+b+1}{2b^2}=0\Rightarrow \frac{(b-1)(4b^2-2b-1)}{2b^2}=0\Rightarrow b=1,4b^2-2b-1=0$
Therefore, we get three values of $b.$
The sum of all values of $b=1+\frac{2}{4}=\frac{3}{2}$