Solve for x.........?

5x - 4 ≥ 0

x ≥ 4/5

x/2 + 2 ≥ 0

x/2 ≥ -2

x ≥ -4

=> x ≥ 4/5

5x - 4 = (x/2 + 2)^2

5x - 4 = x^2/4 + 4 + 2x

20x - 16 = x^2 + 16 + 8x

x^2 - 12x + 32 = 0

d = 144 - 4*32 = 16; sqrt d = 4

x1 = (12+4)/2 = 8

x2 = (12-4)/2 = 4

2√(5x - 4) = x + 4

The expression inside the radical must be positive --> x ≥ 4/5

The RHS must be positive --> x ≥. -4

--> x ≥ 4/5

Square both sides::

4(5x - 4) = (x + 4)^2

20x - 16 = x^2 + 8x + 16

x^2 - 12x + 32 = 0

(x - 4)(x - 8) = 0

x = 4, x = 8

Another method:

10&38730;(5x - 4) = 5x + 20

Substitute t = √(5x - 4) (t ≥ 0)

10t = t^2 + 24

t^2 - 10t + 24 = 0

(t - 4)(t - 6) = 0

t = 4 --> 5x - 4 = 16 --> x = 4

t = 6 --> 5x - 4 = 36 --> x = 8

There are 2 solutions.

the value of x is either 8 or 4

this problem can be solved using the following steps:

1.take the sqrt to the other side of the equation. <it becomes a sq>

2.then simplify it

3. next solve the quadratic eqation

hope this helps!!! :)

sqrt(5x − 4) = (x/2) + 2

2sqrt(5x − 4) = (x + 4)

2(5x - 4) = (x + 4)^2

10x - 8 = x^2 + 8x + 16

I'll spare you the working out, and just tell you that my result was x = 4.

The process was to expand and simplify both sides, until you're free of square roots and fractions. Then you transpose until you get the unknown (x) on one side on its own.

Square both sides and expand the RHS and you can solve it already . Remember to check for extraneous solutions which do not satisfy the equation ,

(5x-4)^2=x/2 +2

25x^2-40x+16=x/2+2

--------------------------------- x 2

50x^2-80x+32=x+4

50x^2-80x+32-x-4=0

50x^2-81x+28=0

50x^2-25x-56x+28=0

25x(2x-1)-28(2x-1)=0

(25x-28)(2x-1) =0

25x-28 =o

25x =28

X1 = 28/25 = 1 3/25

2x-1 =0

2x =1

X2 = 1/2