Question

If $\alpha ,\beta$ are roots of the equation ${\left(3x+2\right)}^2+p\left(3x+2\right)+q=0$, then roots of $x^2+px+q=0$ are-

• A. $\alpha ,\beta$
• B. $3\alpha +2,3\beta +2$
• C. $\frac{1}{3}\left(\alpha -2\right),\frac{1}{3}\left(\beta -2\right)$
• D. $\alpha -2,\beta -2$

Answer 1

The correct option is B $3\alpha +2,3\beta +2$

Given

$(3x+2{)}^2+p(3x+2)+q=\mathrm{0.........}(1)$

Let $3x+2=X$

Now the equation will become

$X^2+pX+q=\mathrm{0............}\left(2\right)$

If $\alpha ,\beta$ are the roots of (1) then roots of (2) will be

$X=3\alpha +2$ and $X=3\beta +2$ [$\because$ both the equations are same the only change is in the value of X]

Now we have to find roots of equation

$x^2+px+q=\mathrm{0............}\left(3\right)$

As we see equation (3) and (2) are symmetric then there roots will be same which means roots of equation (3) are

$x=3\alpha +2,3\beta +2$