Why does (x-2)^2+9=0 have no real solutions?

Mathematics

1 answer:

alex41 [277]2 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Given

(x - 2)² + 9 = 0 ( subtract 9 from both sides )

(x - 2)² = - 9 ( take the square root of both sides )

x - 2 = ± $\sqrt{-9}$

$\sqrt{-9}$ yields no real solutions

There is no real number when multiplied by itself gives - 9

In fact the solutions to $\sqrt{-9}$ are said to be imaginary, that is ± 3i

where i = $\sqrt{-1}$

If interested investigate further.

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