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.
Can someone help me solve this please?
[SQRT(2x2 + 6x + 4)] = x + 1
6 answers
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
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
I like that 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.
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.
0 answers