Question

If sinϴ and cosϴ roots of the quadratic equation  ax²+bx+c=0,prove that a²-b²+2ac=0

Answer 1

Solution: taking theta as q

Comments

Another approach,

we have, sinq+cosq=-b/a...(i)

and sinq*cosq=c/a...(ii)

squaring (i),

sin^2q+2sinq*cosq+cos^2q=(b^2)/(a^2)

or, 1+2*(c/a)=(b/a)^2

or, a^2+2ac=b^2

Now, LHS=a^2-b^2+2ac

=a^2+2ac-b^2=b^2-b^2=0=RHS