Math
posted by Maria .
solve: xsquarerootx2=4
and
write 1+4i over 1i in the form a+bi

#1.
I'm not sure if you mean x  sqrt(x2) = 4 or x  sqrt(x)  2 = 4, but I'll go with the first.
x  sqrt(x2) = 4
To do this, we isolate the term with squareroot to one side of equation:
x  4 = sqrt(x2)
Then we square both sides and solve for x:
(x  4)^2 = (sqrt(x2))^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(62) = 4
6  sqrt(4) = 4
6  2 = 4
4 = 4
Thus x is indeed equal to 6.
x = 3 :
3  sqrt(32) = 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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