# Math

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solve: x-squarerootx-2=4
and
write 1+4i over 1-i in the form a+bi

• Math -

#1.
I'm not sure if you mean x - sqrt(x-2) = 4 or x - sqrt(x) - 2 = 4, but I'll go with the first.
x - sqrt(x-2) = 4
To do this, we isolate the term with squareroot to one side of equation:
x - 4 = sqrt(x-2)
Then we square both sides and solve for x:
(x - 4)^2 = (sqrt(x-2))^2
x^2 - 8x + 16 = x - 2
x^2 - 8x - x + 16 + 2 = 0
x^2 - 9x + 18 = 0
(x - 6)(x - 3) = 0
x = 6 and x = 3
We need to check these values of x by substituting them back to the original:
x = 6 :
6 - sqrt(6-2) = 4
6 - sqrt(4) = 4
6 - 2 = 4
4 = 4
Thus x is indeed equal to 6.

x = 3 :
3 - sqrt(3-2) = 4
3 - sqrt(1) = 4
3 - 1 = 4
2 =
Thus x is NOT equal to 3.

#2.
(1 + 4i) / (1 - i)
Multiply both numerator and denominator by the conjugate of denominator:
= (1 + 4i) / (1 - i) * (1 + i)/(1 + i)
= (1 + 4i + i + 4i^2) / (1 - i^2)
Note that i^2 = -1, thus
= (1 + 5i + 4(-1)) / (1 -(-1))
= (1 + 5i - 4) / 2
= -3/2 + (5/2)i

Hope this helps :3