if a+b+c=0, prove that the roots of ax^2+bx+c=0 are rational. Hence, show that the roots of (p+q)x^2-2px+(p-q)=0 are rational

a + b + c = 0 = b = - a - c

= b = -(a + c)

Discriminant of ax² + bx + c is given by

D = b² - 4ac = { - (a + c)²} - 4ac (Putting value of b)

= (a + c)² - 4ac = a² + c² + 2ac - 4ac

= a² + c² - 2ac = (a - c)² ( A perfect square)

Thus roots = (- b ± √D)/2a = {-b ± √(a - c)²}/2a = (-b + a - c/2a) ( A rational number as a, b, c are rational numbers)

Thus roots = rational.

Thus we can conclude that if discriminant is a perfect square then roots are rational.

For (p + q)x² - 2px + (p - q) = 0

Discriminant = (- 2p)² - 4 (p + q)(p - q)

= 4p² - 4(p² - q²) {As (a + b)(a - b) = (a² - b²) }

= 4p² - 4p² + 4q² = 4q²

= (2q)² (A perfect square)

Here again the discriminant is a perfect square and hence the roots must be rational.

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