With the first method you should check the solutions immediately before starting the second method. How I solved it (2x)²=(4*sqrt(x))² 4x²=16x 4x²-16x=0 4x*(x-4)=0 4x=0 or x-4=0 x=0 or x=4 check the possible solutions x=0 is rejected, so x=4 by replacing the values of x in the equation
However, when you get to 4x^2-16x=0, if you divide both sides by x, you get 4x-16=0, only one solution=4….problem of x=0 is eliminated. I’m pretty sure the 0 solution is an extraneous root caused by squaring both sides of equation. When the original equation was presented it was not a quadratic. Even though squaring both sides is an efficient method in this case, the problem is that the original equation only required one solution, but by creating the quadratic, you force another solution, which in this particular case is not a valid solution.
With the first method you should check the solutions immediately before starting the second method.
How I solved it
(2x)²=(4*sqrt(x))²
4x²=16x
4x²-16x=0
4x*(x-4)=0
4x=0 or x-4=0
x=0 or x=4
check the possible solutions
x=0 is rejected, so x=4 by replacing the values of x in the equation
Thanks 👍
However, when you get to 4x^2-16x=0, if you divide both sides by x, you get 4x-16=0, only one solution=4….problem of x=0 is eliminated. I’m pretty sure the 0 solution is an extraneous root caused by squaring both sides of equation. When the original equation was presented it was not a quadratic. Even though squaring both sides is an efficient method in this case, the problem is that the original equation only required one solution, but by creating the quadratic, you force another solution, which in this particular case is not a valid solution.
4, by inspection.
2(4)/Sqrt[4]=4 x=4 It’s in my head.
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(2x/sqrt(x))² = 4²
4x²/x = 16
4x = 16
x = 4
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X = 0 can't be a valid solution here as x being in the denominator in the original equation.
Решил в уме за три секунды что х=4.
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