Question

If $a$ and $b$ are rational and $b$ is not perfect square, then the quadratic equation with rational coefficients whose one root is $3a+\sqrt{b}$ is

• A. $x^2-6ax+9a^2-b=0$
• B. $3a{x}^2+x-\sqrt{b}=0$
• C. $x^2+3ax+\sqrt{b}=0$
• D. $\sqrt{b}{x}^2+x-3a=0$

Answer 1

Answer: (A)

$x^2-6ax+9a^2-b=0$

Roots of quadratic equation always occur in pairs.

So, if $3a+\sqrt{b}$ is root then $3a-\sqrt{b}$ is also a root.

Let $x^2+px+q$ be quadratic equation

Sum of roots is $3a+\sqrt{b}+3a-\sqrt{b}=6a=-p$

Product of root is $(3a+\sqrt{b})(3a-\sqrt{b})=9a^2-b=q$

So, the quadratic equation will be $x^2-6ax+9a^2-b=0$