(x-3)^4=16 Taking square root on both sides, (x-3)^2=+4 or (x--3)^2=-4 If (x-3)^2=4 taking square root again, x-3=+2 or x-3= -2 x=3+2=5 or x=3-2=1 If (x-3)^2=-4 taking square root x-3=+ or - 2i If x-3= 2i then.x=3+2i If x-3 = -2i then x=3-2i So the solutions are 5,3,3+2i and 3-2i
5 and 1 is the he answer. Actually got it in seconds 16 is 4^2 which is the same as 2^4 or (-2)^4 then cancelling out 4 at the exponent at both sides x-3=2 or x-3=-2 and finally x=5 and 1
@caleboji4857 You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
Observing that the 4th root of 16 is +/-2, I found the real roots, 1 and 5, by inspection. Finding the remaining 3 roots requires a considerable amount of extra work, which the video demonstrated.
Many thanks. Have a nice day. NB. I neglected to state that the 4th root of 16 is +/- 2, not just 2, which I have now corrected. Hence x - 3 = 2, giving x = 5 and x - 3 = - 2, giving x= 1. That is how I was able to obtain the real roots by inspection.
@psycholoogdrs.hansjacobs9816 You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
@harrymatabal8448 You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
@ShimelesTeacher You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
The solution is too long. The shortest solution is to take the 4th root of both sides. Hence, X-3=2. X= 2+3=5; X=5. To check whether 5 is the correct value of X, substitute the X value to the original equation, (5-3)^4= 16, 2^4=16, 16=16. Therefore, 5 is the only value of X.
@gregc.mariano9226 You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
Remember, this is a fourth-degree equation. You don't just take the fourth root of both sides. If you do that, you're just gonna have just two values of x, instead of four.
@Onoelo23gf You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation. So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
Very nice. Sometimes, I have trouble getting started, and you always help me with that. From there, Ican solve.
Awesome! I am really happy to hear that.
Sweet 👍
Thank you
or just u=x-3
u^4=16
recall if x^n=m, then x=nthroot(m) e^{2ikpi/m} where k is an integer
u=2e^{ikpi/2}
x-3=+-2, +-2i
x=3+-2, 3+-2i
(x-3)^4=16
Taking square root on both sides,
(x-3)^2=+4 or (x--3)^2=-4
If (x-3)^2=4 taking square root again, x-3=+2 or x-3= -2
x=3+2=5 or x=3-2=1
If (x-3)^2=-4 taking square root
x-3=+ or - 2i
If x-3= 2i then.x=3+2i
If x-3 = -2i then x=3-2i
So the solutions are
5,3,3+2i and 3-2i
Fantabulous. That's a good approach.
5 and 1 is the he answer. Actually got it in seconds
16 is 4^2 which is the same as 2^4 or (-2)^4 then cancelling out 4 at the exponent at both sides x-3=2 or x-3=-2 and finally x=5 and 1
You miss out in imaginary solutions with such a method
@caleboji4857
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
(x-3)^4=16
x=5
5 - 3 = 2
2^4 =16
1;5;3-2i;3+2i
Very correct.
[(x-3)²=y]
y² = 16
|y| = 4 -> y=4 or y=-4
(x-3)² = 4; (x-3)² = -4
[x-3=z]
z²=4; z²=-4
z=2; z=-2; z=2i; z=-2i
x=2+3; x=-2+3; x=2i + 3; x= -2i + 3
x=5; 1; 3+2i; 3-2i
You nailed it, man. Bravo!!!
Observing that the 4th root of 16 is +/-2, I found the real roots, 1 and 5, by inspection.
Finding the remaining 3 roots requires a considerable amount of extra work, which the video demonstrated.
That's a nice observation. I must commend you for that.
Many thanks. Have a nice day.
NB. I neglected to state that the 4th root of 16 is +/- 2, not just 2, which I have now corrected. Hence x - 3 = 2, giving x = 5 and x - 3 = - 2, giving x= 1. That is how I was able to obtain the real roots by inspection.
X=5 or 1 or (+/-) 2i+3
(X-3)4=2to 4
X-3=2. X=5
@psycholoogdrs.hansjacobs9816
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
(X-3)^4= 2^4
X-3 = +-2
X=3+2
Or X=3-2
Nice one.
X also has two other values.
two solutions are obvious: x=5 and x=1
5 and 1
You are right, man. But this is a fourth-degree equation. So, it was supposed to have four roots (values of x).
@harrymatabal8448
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
(5-3)^4 = 16
(x-3)⁴ = 16
x ?
=========
For real solution
(x-3)⁴ = 16
(x-3)⁴ = (±2)⁴
x-3 = ±2
x - 3 = 2 --> x = 5
x - 3 = -2 --> x = 1
x = 1 & 5 7:13
Look at this shorter method.
(x-3)^4 = 2^4
=> x-3 = 2
x = 2+3 = 5
1, 3+2i, 3-2i
@ShimelesTeacher
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
X-3=+/-2
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
6
Bro it could be done in seconds
I wouldn't disagree.
The solution is too long.
The shortest solution is to take the 4th root of both sides.
Hence, X-3=2. X= 2+3=5; X=5.
To check whether 5 is the correct value of X, substitute the X value to the original equation, (5-3)^4= 16, 2^4=16, 16=16.
Therefore, 5 is the only value of X.
@gregc.mariano9226
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
Way too over-complicated approach. And definitely not an Maths Olympiad question.
I would appreciate it if you would show me your approach.
Rewrite 16 as 2^4 and raise both sides to power 1/4; you will get: x-3 = 2
Too long. 4th root of LHS=±2; Hence x=1 or 5. i is not relevant here.
Remember, this is a fourth-degree equation.
You don't just take the fourth root of both sides. If you do that, you're just gonna have just two values of x, instead of four.
@Onoelo23gf
You always should remember that any power of x tells you how many solutions should be. In this particular case (x^4) there are 4 roots of a given equation.
So we have 2 real solutions and 2 complex solutions. If you have the math problem where the domain of x is not specified, you must show all solutions. If the domain is x = R you can avoid the complex solutions.
X=5