Question

the condition of the roots of ax2 +b= 0 being real , unequal and opposite in sign is-
What???
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Answer 1

Step-by-step explanation:

in photo the solution is there

Answer 2

Answer:

let the zeroes be p, q

For p and q to be real. and unequal,

b² - 4ac > 0

Here, b =0, and c = b

So, -4ab > 0

For p and q to have opposite signs,

p + q = 0

As sum of zeroes = -b/a,

-0/a = 0

As this is true for every value of a,

The final condition is : - 4ab > 0