√[3x + √(3x + 4)] = 4?

Question

does this have a solution

Answer 1

Yes, it does.

As sweet n simple figured, x = 4, 7. Unlike the next answer concludes, both are solutions to the original. (Both ALWAYS are in this situation — people always forget a square root of a positive number can be +/-. They then find one root leading to a problem because they fail to consider the result of a square root operation being negative. In this case, 7 leads to √25 and, if you take the negative result, the outer square root is again of 16, not of 21.)

Finally, the respondent with x = 3/4 just read the original's grouping incorrectly and solved the wrong problem.

So yes, it has a solution. With two results.

Answer 2

Hey there!

Here's the answer.

sqrt(3x+sqrt(3x+4))=4 --> Write the problem.
3x+sqrt(3x+4)=16 --> Square both sides of the equation.
sqrt(3x+4)=-3x+16 --> Subtract 3x from both sides of the equation.
3x+4=(-3x+16)^2 --> Square both sides of the equation.
3x+4=(a+b)^2 --> Substitute a for -3x and b for 16.
3x+4=a^2+2ab+b^2 --> By the formula (a+b)^2= a^2+2ab+b^2.
3x+4=(-3x)^2+2(-3x)(16)+(16)^2 --> Substitute -3x for a and 16 for b.
3x+4=9x^2-96x+256 --> Simplify the right side of the equation.
0=9x^2-99x+252 --> Subtract 3x+4 in both sides of the equation.
9x^2-99x+252=0 --> Switch the parts of the equation around.
9(x^2-11x+28)=0 --> Factor out the GCF i.e. 9.
x^2-11x+28=0 --> Divide 9 in both sides of the equation.
(x-4)(x-7)=0 --> Use Trial and Error to factor out the above expression.
x-4=0, x-7=0 --> Use the zero product property i.e. if pq=0, then p=0 or q=0.
x=4, x=7 --> Solve the equation in terms of x.

The solutions are x=4 or x=7. Check the solutions, by substituting the values for x.

sqrt(3(4)+sqrt(3(4)+4))=4 -->
sqrt(12+sqrt(12+4))=4 -->
sqrt(12+sqrt(16))=4 -->
sqrt(12+4)=4 -->
sqrt(16)=4 -->
4=4, true.

The solution 4 satisfies the equation.

sqrt(3(7)+sqrt(3(7)+4))=4 -->
sqrt(21+sqrt(21+4))=4 -->
sqrt(21+sqrt(25))=4 -->
sqrt(21+5)=4 -->
sqrt(26)=4, false.

Since 7 does not satisfy the solution, this solution is called as an extraneous solution.

Thus the solution to the equation is 4.

Hope it helps!

Answer 3

srqt[ 3x + sqrt(3x + 4) ] = 4

square both sides of the equation

3x + sqrt(3x + 4) = 16

sqrt(3x + 4) = 16 - 3x

square both sides of the equation

3x + 4 = 256 - 96x + 9x^2

0 = 9x^2 - 99x + 252

0 = 9(x^2 - 11x + 28)

0 = 9 (x - 7)(x - 4)

x - 7 = 0 or x - 4 = 0

x = 7 or x = 4

x = 7 is not a solution of the original equation.

Answer: x = 4

Answer 4

3x + √(3x + 4) = 16
√(3x + 4) = 16 - 3x
3x + 4 = 256 - 96x + 9x²
0 = 9x² - 99x + 252
x² - 11x + 28 =0
(x - 7).(x - 4) = 0
x = 4 , x = 7
Test x = 4
√(12 + 4) = √16 = 4 as required
Test x = 7
√21 + 5 ≠ 4
Answer x = 4

Answer 5

√[3x + √(3x + 4)] = 4
squaring both sides
3x + √(3x + 4) = 16
√(3x + 4) =16-3x
again squaring both sides
3x+4 = (16-3x)^2
3x+4 = 256 +9x^2 -96x
9x^2 -99x +252 =0
x^2 - 11x +28=0
(x-7)(x-4)=0
so x= 7 or 4

Answer 6

sqrt(3x+4)+sqrt(3x)=4.............................(1)
(3x+4-(3x=4.................................(2)
divide (2 by (1
sqrt(3x+4)-sqrt(3x=1..........(3))
add (1 and (3))
2sqrt(3x+4)=5
squaring both sides:
4(3x+4)=25
12x=9
x=3/4 answer