Question

How do you find the roots, real and imaginary, of `y= 2(x+1)^2+(-x-2)^2` using the quadratic formula?

Answer 1

Zeros are complex conjugate numbers `-4/3-isqrt2/3` and `-4/3+isqrt2/3`

Explanation:

Let us first expand RHS using identity `(a+b)^2=a^2+2ab+b^2`

`y=2(x+1)^2+(-x-2)^2`

= `2(x^2+2×x×1+1^2)+(-x)^2+2×(-x)×(-2)+(-2)^2`

= `2x^2+4x+2+x^2+4x+4`

= `3x^2+8x+6`

Now zeros of a quadratic function are those values of variable, here `x`, who make the value of function `0` i.e. `3x^2+8x+6=0`. Quadratic formula gives the solution of `ax^2+bx+c=0` as `x=(-b+-sqrt(b^2-4ac))/(2a)`.

Hence, zeros of `3x^2+8x+6` are `x=(-8+-sqrt(8^2-4×3×6))/(2×3)` or

`x=(-8+-sqrt(64-72))/6`

= `(-8+-sqrt(-8))/6`

= `-8/6+-i(2sqrt2)/6`

= `-4/3+-isqrt2/3`.

Hence zeros are complex conjugate numbers `-4/3-isqrt2/3` and `-4/3+isqrt2/3`