Question

If a, b are zeroes of the quadratic polynomial ax2+bx+c then find the value of a2 - b2

Answer 1

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

Answer 2

$\huge\mathtt{\fbox{\red{Answer}}}$

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