Solve. If the equation has no real solution, say so.
Square root 3x+4 =x
Is it √(3x+4) or √(3x) + 4 ?
If first version
√(3x+4) = x
square both sides
3x+4 = x^2
x^2 - 3x - 4 = 0
(x-4)(x+1) = 0
x = 4 or x = -1
I pressed "post answer" too soon
Of course since we squared both sides, extraneous roots could have been introduced, so all answers have to be verified.
if x = 4
LS = √(12+4)
= √16 = 4 = RS
if x = -1
LS = √1 = 1
RS = -1
so x = 4 is the only solution.
To solve the equation square root of 3x + 4 = x, we need to isolate the square root term.
First, let's square both sides of the equation to eliminate the square root:
(sqrt(3x + 4))^2 = x^2
Simplifying the left side of the equation:
3x + 4 = x^2
Now we have a quadratic equation, so let's rearrange it into standard form:
x^2 - 3x - 4 = 0
To solve this quadratic equation, we can use factoring or the quadratic formula. Let's try factoring first:
(x - 4)(x + 1) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 1 = 0
Solving for x in each equation:
x = 4 or x = -1
So the equation has two potential solutions: x = 4 and x = -1.
However, we need to check if these solutions satisfy the original equation. Let's substitute each value into the original equation and see if the equation holds true:
For x = 4:
√(3(4) + 4) = 4
√(12 + 4) = 4
√16 = 4
4 = 4
The equation holds true for x = 4, so it is a valid solution.
For x = -1:
√(3(-1) + 4) = -1
√(-3 + 4) = -1
√1 = -1
1 = -1
The equation does not hold true for x = -1, so it is not a valid solution.
Therefore, the only solution to the equation square root of 3x + 4 = x is x = 4.