Prove that this equation has two rational roots if a does not = c
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a(x^2 +x) + c(x+1)=0

=> ax^2 + ax + cx +c =0

=> ax^2 + (a+c)x +c = 0

So discriminant D = (a+c)^2 - 4ac

=> D = a^2 + 2ac + c^2 -4ac

=> D= a^2 - 2ac + c^2

=> D = (a-c)^2

The square confirms that the roots are real and rational. Since (a-c)^2 is always greater than or equal to zero and is always a perfect square(making it rational).

But if a = c then,

D = (a-c)^2 = (a-a)^2 = 0

And we know when D=0 it has one and only one real root.

Therefore, to satisfy the condition D>0 ie having 2 real roots a must not equal to c.





by Level 7 User (25.7k points)

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