Question

y²= ax²+2bx+c then prove d²x/dy²= (b²-ac)/(ax+b)³

Answer 1

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The roots are complex conjugate

Explanation Is here

If the roots of

ax2+2bx+c=0

are real and distint then b2−ac>0

Now grouping

(a+c)(ax2+2bx+c)−2(ac−b2)(x2+1)=0

we have

(a2+2b2−ac)x2+2b(a+c)x+2b2+c(c−a)=0

and solving for x

x=−b(a+c)+√(4b2+(a−c)2)(ac−b2)a2+2b2−ac

and ac−b2<0 so the roots are complex conjugate