how do i find all the real solutions of an equation?

Question

the equation is {x minus the square root of 4x minus 4 equals 0}

Answer 1

Hi Brittany;

I MADE A CORRECTION.  b OF THE QUADRATIC FORMULA SHOULD BE -b.  SORRY.

I think this is...

x-√(4x)-4=0

Please correct me if I am wrong...

Let's isolate the √.

x-4=√(4x)

Let's square both sides...

(x-4)²=[√(4x)]²

x²-8x+16=4x

Let's subtract 4x from both sides...

x²-12x+16=0

Let's use the quadratic formula...

x=[-b+/-√(b²-4ac)]/2a

x={12+/-√[-12²-(4*1*16)]}/[(2)(1)]

x=[12+/-√(144-64)]}/2

x=(12+/-√80)/2

x={12+/-√[(16)(5)]}/2

x=(12+/-4√5)/2

x=6+/-2√5

Answer 2

Vivian worked it correctly if the radical is only around the 4x.  I think I've seen this problem before is some math book and the radical goes around the (4x-4) so the problem looks like

x-√(4x-4)=0

 

If that is case, simply square both sides:

 

x²-4x-4=0

 

Then factor using the rule (a-b)²=a²-2ab+b²

 

(x-2)²=0

 

Take the square root

x-2=0

x=2