I need some math help?

Multiply and simplify the following radical expressions.
(2 root 3 √4 - 3 root 3 √2)(3 root 3 √4 + 2 root 3 √10)

( √3x + √3)2nd power

Rationalize the denominator
7/ √24b3

25/root 4 √8a

√7 - √2/ √2 + √7

√x - √y/ √x + √y

Solve

√3t + 7 - t = 1

Divide the following complex number write the answer in the form a + bi

7 + 3i/4 - 2i

simplify i97

for the solve section, the square root thing is over the 3t + 7.

cidyah2016-01-28T06:40:07Z

(√7 - √2)/ (√2 + √7 )

Multiply and divide by (√7+√2)
= (√7-√2)(√7+√2) / ((√7+√2)(√7+√2))

The numerator is of the form (a-b)(a+b) = a^2-b^2
a= √7
a^2= 7
b=√2
b^2 = 2
a^2-b^2 = 7-2 = 5

= 5 / ((√7+√2)(√7+√2))
= 5 / (7+√14+√14+2)
= 5 / (9+2√14)

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(7 + 3i)/ (4 - 2i )

Multiply and divide by (4+2i)
(7+3i)(4+2i) / ((4-2i)(4+2i))

The denominator is of the form (a-b)(a+b) = a^2-b^2
a=4
b= 2i
a^2=16
b^2= 4i^2 = 4(-1) = -4
a^2-b^2 = 16-(-4) = 20

= (7+3i)(4+2i) / 20
= (28+12i+14i+6i^2) / 20
= (28+26i-6) / 20
= (22+26i) / 20
= 22/20 + (26/20) i
= 11/10 + (13/10)i

(7 + 3i)/ (4 - 2i ) = 11/10 + (13/10)i

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√3t + 7 - t = 1
sqrt( 3t+7) - t = 1
sqrt(3t+7) = t+1
square both sides
3t+7 = (t+1)^2
3t+7 = t^2+2t+1
t^2+2t+1 -3t-7 = 0
t^2-t-6 = 0
(t-3)(t+2) = 0
t = 3 or t = -2

t=-2 does not satisfy √3t + 7 - t = 1
t= 3