px^2+(p+q)x+q=0 show that p=q
kvnmurty:
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.
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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.
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.
Answered by
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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
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
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