Asked by Lucas
Can someone help me solve this please?
[SQRT(2x2 + 6x + 4)] = x + 1
[SQRT(2x2 + 6x + 4)] = x + 1
Answers
Answered by
drwls
Square both sides.
2x^2 + 6x + 4 = x^2 +2x + 1
x^2 + 4x + 3 = 0
(x + 3)(x + 1) = 0
x = -3 or -1
The -3 solution assumes -2 is a valid square root of root 4. Some teachers may disagree.
2x^2 + 6x + 4 = x^2 +2x + 1
x^2 + 4x + 3 = 0
(x + 3)(x + 1) = 0
x = -3 or -1
The -3 solution assumes -2 is a valid square root of root 4. Some teachers may disagree.
Answered by
Damon
2 x^2 + 6 x + 4 = x^2 + 2 x + 1
x^2 + 4 x + 3 = 0
(x+3)(x+1) = 0
x = -3 or x = -1 only one may work so check
x = -3
sqrt (18- 18+4) = -3+1
sqrt 4 = -2 ? well, not really
check
x = -1
sqrt (2 -6+4) = -1+1 = 0 yes
x^2 + 4 x + 3 = 0
(x+3)(x+1) = 0
x = -3 or x = -1 only one may work so check
x = -3
sqrt (18- 18+4) = -3+1
sqrt 4 = -2 ? well, not really
check
x = -1
sqrt (2 -6+4) = -1+1 = 0 yes
Answered by
Count Iblis
Square both sides:
2x^2 + 6x + 4 = (x+1)^2
We also have to demand that the argument of the square root is positive. So,
2x^2 + 6x + 4 >=0,
but any solution of the above quadratic equation will automatically satisfy this condition, because the right hand side is a perfect square (x+1)^2 which is always non-negative.
So, we have to solve the equation:
x^2 + 4 x + 3 = 0 -->
(x+1)(x+3) = 0 --->
x = -1, or x = -3
2x^2 + 6x + 4 = (x+1)^2
We also have to demand that the argument of the square root is positive. So,
2x^2 + 6x + 4 >=0,
but any solution of the above quadratic equation will automatically satisfy this condition, because the right hand side is a perfect square (x+1)^2 which is always non-negative.
So, we have to solve the equation:
x^2 + 4 x + 3 = 0 -->
(x+1)(x+3) = 0 --->
x = -1, or x = -3
Answered by
Damon
I like that Count Iblis :)
Answered by
Count Iblis
I forgot the condition that the square root itself must be positve as pointed out by Drwls and Damon. So, only x = -1 is a valid solution.
Note that you could define a different square root function that is always less or equal than zero, but in that case, you have only the x = -3 as the solution.
Note that you could define a different square root function that is always less or equal than zero, but in that case, you have only the x = -3 as the solution.
Answered by
Damon
We discussed this the other day and decided that for working with square root functions we should hedge the subject to saying sqrt of anything is +, but the negative is perfectly reasonable.