kellyrosee kellyrosee
  • 22-03-2024
  • Mathematics
contestada

Solving quadratics with imaginary solutions

Solving quadratics with imaginary solutions class=

Respuesta :

lirich lirich
  • 22-03-2024

Answer:

x = (1+2isqrt6)/5, (1-2isqrt6)/5

Step-by-step explanation:

10x^2-4x+10=0

a=10, b=-4, c=10

by quadratic formula: (you forgot to negate b)

x = (- (-4) +- sqrt(16-4*10*10))/(2*10)

= (4+-sqrt(-384))/20

Here we are simplifying the radical:

= (4+- (i)sqrt(64*6))/20

That was because sqrt(-n) is the same as (i * sqrt(n))

= (4+- (i)sqrt(8^2*6))/20

= (4+- (8i)sqrt(6))/20

= (1+-2isqrt6)/5

Two solutions to +- expression:

x = (1 + 2isqrt6)/5, (1-2isqrt6)/5

Hope this helps

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