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

1 Answer
Shwetank Mauria
Jul 20, 2016

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

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