Question

If the zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal, then (A) c and a have opposite signs (B) c and b have opposite signs (C) c and a have the same sign (D) c and b have the same sign

Answer 1

Answer:

option A. c and a have + sign

Answer 2

<<solution>>

ax²+bx+c=0

x²+(b/a)X+(c/a)=0

let the zeroes are p and q...

therefore

p+q=-b/a

pq=c/a

now.....

(p-q)²=(p+q)²-4pq

=>p-q=√[(b²/a²)-4c/a]

now ....

p+q=b/a

p-q=√[(b²/a²)-4c/a]

____________________

2p=(b/a)+√[(b²/a²)-4c/a]

$p = \frac{ \frac{b}{a} + \sqrt{ \frac{b {}^{2} }{a {}^{2} } - \frac{4c}{a} } }{2}$

and ...

$q= \frac{ \frac{b}{a} - \sqrt{ \frac{b {}^{2} }{a {}^{2} } - \frac{4c}{a} } }{2}$

now...

$p = q \\ = > \frac{ \frac{b}{a} + \sqrt{ \frac{b {}^{2} }{a {}^{2} } - \frac{4c}{a} } }{2} = \frac{ \frac{b}{a} - \sqrt{ \frac{b {}^{2} }{a {}^{2} } - \frac{4c}{a} } }{2} \\ = > \frac{b {}^{2} }{a {}^{2} } = \frac{4c}{a} \\ = > b {}^{2} = 4ac$

option (c) c and a have the same sign