SOLVE: �ã(4x+ 5) + 1 = 0
�ã(4x+ 5)= -1
(�ã(4x+ 5))squared= (-1)squared
4x + 5 = 1
I know I solve for x
I get that x = 1
but my book states that x = -1
Please show me how this is true
I don't understand what the term �ã means. Can you retype the problem some other way?
Your third line suggests to me that the original question was
√(4x+5) + 1 = 0
from your "4x + 5 = 1 "
4x = 1 - 5
4x=-4
x=-1 as your book states.
HOWEVER, as soon as you reached your second line of
√(4x+5) = -1 you could have stopped, since that contradicts the definition of the square root symbol, which is defined as the positive square root of a number.
eg. suppose you had x^2 = 9
Many students are taught to have as their next line
x= ± 3 whereas the actual should have been ± x = 3
For equations it really does not matter but in inequations such as
x^2 > 9 it is actually WRONG to say
x > ± 3 and you must have ± x > 3
Secondly, once you square both sides of an equation like you did in your third line, all answers you obtain must be verified in the original equation.
So in this case the correct answer would be
"no solution in the real number set"
I apologize for the confusion caused by the symbol �ã in the original question. Based on your explanation, it seems that the correct interpretation of the equation is √(4x+5) + 1 = 0.
To solve this equation, we can follow these steps:
1. Subtract 1 from both sides of the equation: √(4x+5) = -1.
2. As you correctly pointed out, this equation contradicts the definition of the square root because the square root of any number is always positive. Therefore, this equation has no solution in the real number set.
When you squared both sides of the equation (√(4x+5))^2 = (-1)^2, you introduced an extraneous solution. This means that you obtained a solution that does not satisfy the original equation. In this case, squaring both sides led to an incorrect solution because the original equation had no solution in the first place.
It is important to always check the solutions obtained by squaring both sides of an equation to ensure they are valid in the original equation. In this case, we can see that the solution x = -1, as stated in your book, is not valid because plugging it back into the original equation would result in √(4(-1)+5) + 1 = 0 becoming √(1) + 1 = 0, which is not true.