Question

twice the sum of the square of a number and 1, increased by the square of the quantity 1 more than the square of the number gives 3. Find the numbers. Answers are 0,+- 2i

Answer 1

2(x^2+1)+(1+x^2)^2=3

Answer 2

Let 1+x^2=t
So. t^2+2t=3
t^2+2t-3=0
You will get t=1,-3
Put 1+x^2=1
x^2=1-1=0
So x=0
Now put 1+x^2=-3
x^2=-2
x=+-2i
So we get three values of x 0,+2i,-2i

Answer 3

Any appreciation for this help

Answer 4

\[2(x^2+1)+(x^2+1)^2=3\]
\[2x^2+2+x^4+2x^2+1=3\]\[x^4+4x^2+3=3\]\[x^4+4x^2=0\]\[x^2(x^2+4)=0\]\[x^2=0\]that gives us our first answer of \(x = 0\)\[x^2+4=0\]\[x^2=-4\]\[\sqrt{x^2} = \sqrt{-4}\]\[\sqrt{x^2} = \sqrt{4*-1}\]\[\sqrt{x^2} = 2\sqrt{-1}\]\[x=\pm2i\]for our other two answers

Answer 5

@pawanyadav you made a small algebra error:

Now put 1+x^2=-3
x^2=-2
x=+-2i

that should be \[1+x^2=-3\]\[1-1+x^2 = -3-1\]\[x^2=-4\]\[x = \pm 2i\]

Answer 6

Sorry wrongly typed