Question

If equation x²-3x+1=p(x-3) has equal roots find the possible values of p

Answer 1

Answer:

The answer is $p=1,5$

Step-by-step explanation:

$x^2-3x+1=p(x-3)\\\Rightarrow x^2-3x+1=px-3p\\\Rightarrow x^2-3x-px+3p+1=0\\\Rightarrow x^2-(3+p)x+(3p+1)=0$

Now for equal roots we must have

      $b^2-4ac=0$

$\Rightarrow (-(3+p))^2-4(1)(3p+1)=0\\\Rightarrow p^2+6p+9-12p-4=0\\\Rightarrow p^2-6p+5=0\\\Rightarrow (p-1)(p-5)=0\\\Rightarrow p=1,5$