# Px^2+(p+q)x+q=0 show that p=q

2
by shanmukhkumar
p-q)^2 is always > 0 for any real values of p and q , as a square of a number (p-q) is always >=0... this does not mean that p = q.

2015-01-15T18:14:00+05:30

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The equation editor is not properly working so i write in the normal text.

b^2- 4 ac = (p+q)^2 - 4 pq = (p-q)^2

If the roots of the given equation  both are equal, then the discriminant is 0.
then (p-q)^2 = 0.  Hence  p = q.

if p= q, then  equation will be x^+2x+1=0,  hence roots are equal and = -1.

If  p > q and  q > p  -- in these case real roots exist  but they are not equal.

2015-01-15T19:10:19+05:30
ATQ
px² + (p+q)x + q = 0
⇒ px² + px + qx  + q = 0
⇒ px(x+1) +q (x +1) = 0
⇒ (px+q)(x+1) = 0
Then we have either px+q= 0 or x+1=0
px +q =0  ⇒ x = -q/p
and x + 1 = 0  ⇒ x = -1
This gives - q/p = -1
⇒  p = q