Question

a(x^2-1)+(a^2+1)x=0 can somebody do it using quadratic formula? Thanks

Answer 1

Answer:

I think the answer is 0

Step-by-step explanation:

Answer 2

Could you check your question again?

If your question was $a(x^2-1)+(a^2+1)x=0$

→ $ax^2+(a^2+1)x-a=0$

→ $\boxed{x=\frac{-(a^2+1)\pm\sqrt{(a^2+1)^2+4a^2} }{2a} }$

→ $\boxed{x=\frac{-a^2-1\pm\sqrt{a^4+6a^2+1} }{2a} }$

If your question was $a(x^2+1)+(a^2+1)x=0$

→ $ax^2+(a^2+1)x+a=0$

→ $\boxed{x=\frac{-(a^2+1)\pm\sqrt{(a^2+1)^2-4a^2} }{2a} }$

→ $\boxed{x=\frac{-(a^2+1)\pm\sqrt{(a^2-1)^2} }{2a} }$

→ $\boxed{x=\frac{-(a^2+1)\pm(a^2-1)}{2a} }$

→ Which is, x=-a or x=-1/a.