Question

The equation is : px^2+qx+r=0
and 'a' is the repeated root.
f(x)=px^2+qx+r=0
f(a)=p(x-a)(x-a)

HOW DID THEY ARRIVE AT THIS?

Answer 1

my dear then notice in the question that a is repeated root so u should make two equation and compare them with each other. coefficient of x with same degree should be equal . consider what you have .
f(x)=px^2+qx+r=0
f(a)=p(x-a)(x-a)
first equation is : f(x)=px^2+qx+r=0
and the second one is : f(a)=p(x-a)(x-a) =px^2+2ax+a^2=0
we suppose to compare coefficient of x^2 , they should be equal and they are .
for the second part compare q with 2a so we find out that q is equal with 2a :q=2a
and with the same way we find that :a^2=r

Answer 2

thank you

Answer 3

ur welcome dear