Question

If $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0,a\ne 0$. Find the equation with roots $k\alpha +s\&k\beta +s,$ where $k\&s$ are constants and $k\ne 0.$

• A. None of the above
• B. $a{\left(\frac{x+s}{k}\right)}^2-b\left(\frac{x+s}{k}\right)+c=0$
• C. $a{\left(\frac{x-s}{k}\right)}^2+b\left(\frac{x-s}{k}\right)+c=0$
• D. $a{\left(\frac{x+s}{k}\right)}^2+b\left(\frac{x+s}{k}\right)+c=0$

Answer 1

The correct option is C $a{\left(\frac{x-s}{k}\right)}^2+b\left(\frac{x-s}{k}\right)+c=0$

Given: $a{x}^2+bx+c=0,a\ne 0$ with roots $\alpha ,\beta$
To find: Transformed equation with roots as $k\alpha +s\&k\beta +s$

Let $t=kx+s$

$\because x=\alpha \Rightarrow t=k\alpha +s$
$\&x=\beta \Rightarrow t=k\beta +s$

Replacing $x\text{with}\frac{t-s}{k},$ we get the equation as:

$a{\left(\frac{t-s}{k}\right)}^2+b\left(\frac{t-s}{k}\right)+c=0$

In terms of variable $x,$ the transformed equation will be:
$a{\left(\frac{x-s}{k}\right)}^2+b\left(\frac{x-s}{k}\right)+c=0$