Question

If one zero of the quadratic polynomial $(a^2+9)x^2+13x+6a$ is reciprocal of the other, then find the value of a.

• A.
  $2$
• B.
  $3$
• C.
  $4$
• D.
  $5$

Answer 1

The correct option is B

$3$
For Polynomial $(a^2+9)x^2+13x+6a,\text{Let the zeroes be}\alpha and\frac{1}{\alpha }\text{Sum of zeroes}=\frac{-b}{a}\Rightarrow \alpha +\frac{1}{\alpha }=\frac{-\left(13\right)}{a^2+9}\text{Product of zeroes}=\frac{c}{a}\Rightarrow \alpha .\frac{1}{\alpha }=\frac{6a}{a^2+9}\Rightarrow 1=\frac{6a}{a^2+9}\Rightarrow a^2+9=6a\Rightarrow a^2-6a+9=0\Rightarrow \left(a^2\right)-2\times a\times 3+(3{)}^2=0\Rightarrow (a-3{)}^2=0\Rightarrow a-3=0\Rightarrow a=3$
Hence, the correct option is b.