If a, b ,c are real numbers such that ac \neq 0, then show that at least one of the equatiions $a x^{2} + b x + c = 0$ and $- a x^{2} + b x + c = 0$

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Solution

Quadratic equation has real roots when Determinant is greater than or equal to 0 (i.e non-negative)

Determinant of 1st equation = b^2 - 4ac

Determinant of 2nd equation is b^2 -4(-a)c = b^2+4ac

If b^2-4ac >= 0 then 1st equation has real roots so meets the condition that one of the equation has real roots

Now suppose the first equation does not have real roots, i.e. b^2-4ac < 0

i.e. b^2 < 4ac

b^2 is always positive (being a perfect square) and therefore 4ac (being greater than b^2) is also always +ve

The determinant of the second equation b^2+4ac is positive (being sum of two +ve numbers)

So the second equation has real roots and so again meets the condition that at least one of the equation has real roots

This proves that at least one of the two given equations has real roots

Hope you get it

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