x2-(p2+q2)x+p2q2=0 Answer it for 100 points
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x^2- (p^2+q^2)x + p^2q^2= 0
=x^2- p^2x- q^2x + p^2q^2= 0
=x(x-p^2)- q^2(x-p^2)= 0
=(x-p^2) (x-q^2)= 0
either x-p^2= 0 or x-q^2=0
x= p^2 x=q^2
=x^2- p^2x- q^2x + p^2q^2= 0
=x(x-p^2)- q^2(x-p^2)= 0
=(x-p^2) (x-q^2)= 0
either x-p^2= 0 or x-q^2=0
x= p^2 x=q^2
delhitian:
Thanks and congrats for points
Answered by
4
x²- (p²+q²)x + p²q² = 0
x² - p²x - q²x + p²q² = 0
x(x - p²) - q²(x - p²) = 0
(x - p²)(x - q²) = 0
=> x - p² = 0
x = p²
=> x - q² = 0
x = q²
Hope it helps
x² - p²x - q²x + p²q² = 0
x(x - p²) - q²(x - p²) = 0
(x - p²)(x - q²) = 0
=> x - p² = 0
x = p²
=> x - q² = 0
x = q²
Hope it helps
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