Posted by **Maria** on Sunday, November 3, 2013 at 10:40pm.

solve: x-squarerootx-2=4

and

write 1+4i over 1-i in the form a+bi

- Math -
**Jai**, Sunday, November 3, 2013 at 10:55pm
#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

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