Question

A and B are the two zeroes of polynomials ax²+bx+c. Find the value of

A²+B².​

Answer 1

Answer:

Let given quadratic polynomial be,

f(x)=ax²+bx+c

Let A and B be the zeroes of f(x).

Then we know that,

A={-b+√(b²-4ac)}/2a

and,

B={-b-√(b²-4ac)}/2a

Now we have,

=A²

=[{-b+√(b²-4ac)}/2a]²

={b²-2b√(b²-4ac)+b²-4ac}/4a²

={2b²-4ac-2b√(b²-4ac)}/4a²

and

=B²

=[{-b-√(b²-4ac)}/2a]²

={b²+2b√(b²-4ac)+b²-4ac}/4a²

={2b²-4ac+2b√(b²-4ac)}/4a²

Now,

=A²-B²

={2b²-4ac-2b√(b²-4ac)}/4a²-{2b²-4ac+2b√(b²-4ac)}/4a²

={2b²-4ac-2b√(b²-4ac)-2b²+4ac-2b√(b²-4ac)}/4a²

={-4b√(b²-4ac)}/4a²

=-b√(b²-4ac)/a²

Hence,

A²-B²=-b√(b²-4ac)/a²

Step-by-step explanation:

This is the answer.

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Answer 2

Step-by-step explanation:

A+B= -b/a

AB= c/a

A²+B²= (A+B)²-2AB

= b²/a² - 2c/a

= (b²-2ac)/a²