Y=px+square root of (4+p^2) is which type of equations
Answers
Answered by
0
Given real non-zero p,q, let a,b be the roots of px2+qx+q=0. Then,
qp=ba+ab+2
I glanced at the other answers to see if there were indeed proofs of what looked to me like an incorrect statement. Of course there were.
Assuming real, non-zero p and q, the roots will be real or a complex conjugate pair, so their quotient (ratio) is real. Let’s make clear a and b are the roots and r=a/b is the ratio.
That makes each square root the square root of a real number, and by convention the radical sign refers to the principal value, a positive real or a positive real times i. So the sum of real square roots cannot be zero unless each individual square root is zero, which isn’t possible here. So presumably the proofs that show this works are wrong.
qp=ba+ab+2
I glanced at the other answers to see if there were indeed proofs of what looked to me like an incorrect statement. Of course there were.
Assuming real, non-zero p and q, the roots will be real or a complex conjugate pair, so their quotient (ratio) is real. Let’s make clear a and b are the roots and r=a/b is the ratio.
That makes each square root the square root of a real number, and by convention the radical sign refers to the principal value, a positive real or a positive real times i. So the sum of real square roots cannot be zero unless each individual square root is zero, which isn’t possible here. So presumably the proofs that show this works are wrong.
Similar questions