Question

If $x^2-8x+15=0$ has roots $p,q$, then which of the following can not be the equation with the roots $p^2,q^2$ ?

• A. $x^2-34x-225=0$
  
• B. $x^2+34x-225=0$
  
• C. $x^2+34x+225=0$
  
• D. $x^2-34x+225=0$
  

Answer 1

Answer: (A), (B), (C)

$x^2+34x+225=0$


Given that $x^2-8x+15=0$ has roots $p,q$.
on comparing with general form $a{x}^2+bx+c=0$, we have
$a=1,b=-8,c=15$

If $a{x}^2+bx+c=0,a\ne 0$ has roots $\alpha ,\beta$
then the transformed quadratic equation with roots ${\alpha }^2$ and ${\beta }^2$ can be written as:
$a(\sqrt{x}{)}^2+b(\sqrt{x})+c=0$

So directly we can say that the required equation is:
$1(\sqrt{x}{)}^2-8(\sqrt{x})+15=0$
$\Rightarrow x-8\left(\sqrt{x}\right)+15=0$
$(x+15{)}^2=(8\left(\sqrt{x}\right){)}^2=64x$
$\Rightarrow x^2-34x+225=0$