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July 28, 2014

July 28, 2014

Posted by **JP** on Thursday, December 11, 2008 at 11:32pm.

Radical (x^2-10x)=3i

2) Simplify

(5+ Radical 2)/(5-Radical 2)

3)describe the nature of the roots

x^2-x-6=0

x^2+4x+29=0

4) Solve for x and check

Radical (x^2 +4x +44) +3=2x

I got 7 and -5/3 as answers, was not sure though

- Math -
**drwls**, Thursday, December 11, 2008 at 11:43pm2) Use the fact that (a + b)(a - b) = a^2 - b^2

3) Let's say your equation in the format ax^2 + bx + c = 0

In your two cases, a = 1.

Calculate the quantity b^2 - 4ac for each equation. If it is positive there are two real roots. If it is zero there is one. If it is negative, there are two complex roots

- Math -
**Reiny**, Friday, December 12, 2008 at 12:10am1) square both sides

x^2 - 10x = 9i^2

x^2 - 10x = -9

x^2 - 10x +9 = 0

(x-1)(x-9) = 0

x = 1 or x=9

check (since we squared)

if x=1

LS = √(1-10) = √-9 = 3i = RS

if x=9

LS = √(81-90) = √-9 = 3i = RS

so x = 1 or x=9

- Math -
**Reiny**, Friday, December 12, 2008 at 12:15am4) since you clearly squared both sides to solve, each answer you obtained has to be verified in the original equation

if x=7 it works

if x= -5/3 it does not work

so x = 7 is the only solution

- Math -
**Reiny**, Friday, December 12, 2008 at 12:18am4) since you clearly squared both sides to solve, each answer you obtained has to be verified in the original equation

if x=7 it works

if x= -5/3 it does not work

so x = 7 is the only solution

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