Solve each equation

-x^2 + 4x = 13      re-arrange

-x^2+4x-13 = 0     easier to work with if you multiply both sides by -1

x^2-4x+13 = 0      Use quadratic Formula   a = 1    b = -4    c = 13

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)      results in      [4+-sqrt(16-4(1)13) ] / 2    =   2 +- sqrt(-36)/2   =   2 +- sqrt 3i

10

-3 + √[10x + 9 ] = x         the idea is to get the radical on one side and everything else on the other

Also....these problems sometimes produce extraneous solutions....we need to be aware of that

Add 3 to both sides

√[10x + 9 ]  = x + 3       square both sides

10x + 9  =  x^2 + 6x + 9    

10x = x^2 + 6x

x^2 - 4x  = 0

x ( x - 4) = 0

x = 0     or   x = 4

Check that both of these solutions are good !!!

b)   Divide both sides by 2   to make the numbers smaller

√ [ 44 - 2x ]  =   x - 10          square both sides

44 - 2x  =    x^2 - 20x + 100     rearrange as

x^2 - 18x + 56 =  0      factor as

(x - 14) ( x - 4) =  0

Set both factors to 0 and solve for x and we get that

x = 14     or  x  = 4

The first solution is good, GM

The second isn't because it makes the right side of the original equation negative.......but we cannot get a negative result from a positive radical....!!!!

Last one...square both sides straight away  and we get that

x^2 - 8x + 16 = 2x     rearrange as

x^2 - 10x + 16  = 0    factor

(x - 8) ( x - 2)  = 0

x = 8      or  x = 2

Note that 8 does not work   because 8 - 4   does not equal     - √ (2 * 8)  =  - √16   =  - 

Note that   x = 2 is good

2 - 4  = - √(2 * 2)

-2  = -√4

-2 = - 2   !!!