4 is the root of 2x-3=6
Answers
Answer:
what's the question write it nicely
Step 1 :
Isolate a square root on the left hand side :
Original equation
√4x-3+√2x+3 = 6
Isolate
√4x-3 = -√2x+3+6
Step 2 :
Eliminate the radical on the left hand side :
Raise both sides to the second power
(√4x-3)2 = (-√2x+3+6)2
After squaring
4x-3 = 2x+3+36-12√2x+3
Step 3 :
Get remaining radical by itself :
Current equation
4x-3 = 2x+3+36-12√2x+3
Isolate radical on the left hand side
12√2x+3 = -4x+3+2x+3+36
Tidy up
12√2x+3 = -2x+42
Step 4 :
Eliminate the radical on the left hand side :
Raise both sides to the second power
(12√2x+3)2 = (-2x+42)2
After squaring
288x+432 = 4x2-168x+1764
Step 5 :
Solve the quadratic equation :
Rearranged equation
4x2 - 456x + 1332 = 0
This equation has two rational roots:
{x1, x2}={111, 3}
Step 6 :
Check that the first solution is correct :
Original equation, root isolated, after tidy up
√4x-3 = -√2x+3+6
Plug in 111 for x
√4•(111)-3 = -√2•(111)+3+6
Simplify
√441 = -9
Solution does not check
21 ≠ -9
Step 7 :
Check that the second solution is correct :
Original equation, root isolated, after tidy up
√4x-3 = -√2x+3+6
Plug in 3 for x
√4•(3)-3 = -√2•(3)+3+6
Simplify
√9 = 3
Solution checks !!
Solution is:
x = 3