What is the answer to j(x)=sqrt(2-x) ?

The answer to j(x)=sqrt(2-x) is x=(-1+sqrt(1+8j^2))/(2j^2) {j 0},x=(-1-sqrt(1+8j^2))/(2j^2)

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j(x)=sqrt(2-x)

j (x) = 2 - x

x = \frac{- 1 + 1 + 8 j ^{2}}{2 j ^{2}} {j < 0 or j > 0}, x = \frac{- 1 - 1 + 8 j ^{2}}{2 j ^{2}}

Solution steps

j (x) = 2 - x

Square both sides:

j^{2} x^{2} = 2 - x

j^{2} x^{2} = 2 - x

Solve

j^{2} x^{2} = 2 - x : x = \frac{- 1 + 1 + 8 j ^{2}}{2 j ^{2}}, x = \frac{- 1 - 1 + 8 j ^{2}}{2 j ^{2}}

x = \frac{- 1 + 1 + 8 j ^{2}}{2 j ^{2}}, x = \frac{- 1 - 1 + 8 j ^{2}}{2 j ^{2}}

Verify Solutions:

x = \frac{- 1 + 1 + 8 j ^{2}}{2 j ^{2}} {j < 0 or j > 0}, x = \frac{- 1 - 1 + 8 j ^{2}}{2 j ^{2}}

True

The solutions are

x = \frac{- 1 + 1 + 8 j ^{2}}{2 j ^{2}} {j < 0 or j > 0}, x = \frac{- 1 - 1 + 8 j ^{2}}{2 j ^{2}}

x^{2} + 3 x + 2 = x + 2

x - 2 - x + 4 = 0

x^{2} - 12 = - x

3 x + 13 = x + 3

What is the answer to j(x)=sqrt(2-x) ?
The answer to j(x)=sqrt(2-x) is x=(-1+sqrt(1+8j^2))/(2j^2) {j<0\lor j>0},x=(-1-sqrt(1+8j^2))/(2j^2)