SOLVE (x - 2)^2 + (x + 1)^2 = 0 MEDAL AWARDED!!!!!
(x - 2)^2 + (x + 1)^2 = 0 Given (x - 2)^2 = -(x + 1)^2 Subtraction Property of Equality \(\dfrac{(x - 2)^2}{(x + 1)^2} = -1\) Division Property of Equality \(\left(\dfrac{x - 2}{x + 1}\right)^2 = -1\) Property of Exponents \(\dfrac{x - 2}{x + 1} = \pm\sqrt{-1}\) Square root property \(\dfrac{x - 2}{x + 1} = \pm i\) Property of imaginary number \(\dfrac{x + 1 - 3}{x + 1} = \pm i\) Property of Expansion of a number \(\dfrac{x + 1}{x + 1} - \dfrac{3}{x + 1} = \pm i\) Property of fractions \(1 - \dfrac{3}{x + 1} =\pm i\) Property of Mixed Fraction
\(1 \pm i = \dfrac{3}{x + 1}\) Addition Property of Equality
\(x + 1 = \dfrac{3}{1 \pm i}\) Multiplication Property of equality
\(x = \dfrac{3}{1 \pm i} - 1\)
i got 1 +- 3i/2
okay thank you
Actually, according to my sources the equivalent result is \(\dfrac{1}{2} \pm \dfrac{3i}{2}\)
so i got it wrong?
wait no thats what i wrote
You wrote 1+- 3i/2
Which is not the same as 1/2 +or - (3i)/2
yah on the computer because i don't know how to write it like that but on my paper i wrote the correct way
Okay, whatever you say... :)
alright thanks…..
@Hero he just forgot the write the parentheses. he meant (1+-3i)/2
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