#1**0 **

-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

ElectricPavlov Feb 22, 2019

#3**+4 **

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 !!!

CPhill Feb 22, 2019

#4**+3 **

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....!!!!

CPhill Feb 22, 2019

#5**+3 **

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 !!!

CPhill Feb 22, 2019