Here's one of the questions I have problems with:
cube root of (x+2)=6th root of (9x+10)
A. -1, 6
B. [-13 +/- sqrt(193)]/2
C. 1, -6
D. [13 +/- sqrt(145)]/2
Please explain how to solve this. Should you cube both sides, or raise both to the 6th? and I get lost trying to do more than square a radical. i'd show my "work" which is basically nonsense but it's useless because I honestly have no idea how to solve this.
thanks in advance for the help. having an example of how to solve these kinds of problems would help me a lot, so if you can step by step would be great.
sorry to post twice; computer meltdown...
(x+2)^(1/3) = (9x+10)^(1/6)
raise both sides to the sixth
[ (x+2)^(1/3) ] ^6 = (x+2)^2
[ (9x+10)^(1/6)]^6 = (9x+10)^1 = 9x+10
so
x^2 + 4 x + 4 = 9 x + 10
x^2 -5 x - 6 = 0
(x-6)(x+1) = 0
x = 6 or x = -1
it really helps to write 15th root of x for example as x^(1/15) power
Oh, and always check the answers
in this case you have for x = -1
1^1/3 = 1^1/6 yes, ok
and for x = 6
8^1/3 = 64^1/6 ? = (8*8)^1/6 =(2*2*2*2*2*2)^1/6 = 2
2 = 2 sure enough
LOL, if I had really noticed it was multiple choice I would have tried x = -1 immediately :)
thanks, Damon - you are awesome =)
To solve the equation cube root of (x+2) = 6th root of (9x+10), we can start by raising both sides to the 6th power. This will help eliminate the roots and simplify the equation.
Step 1: Raise both sides to the 6th power: [(cube root of (x+2)]^6 = [(6th root of (9x+10)]^6
Step 2: Simplify both sides:
(x+2)^6 = (9x+10)^1
Step 3: Expand both sides using the binomial theorem. On the left side, we can use (a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6. On the right side, since 9x+10 is to the power of 1, it remains the same.
(x^6 + 6x^5(2) + 15x^4(2)^2 + 20x^3(2)^3 + 15x^2(2)^4 + 6x(2)^5 + 2^6) = 9x + 10
Simplifying further:
x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64 = 9x + 10
Step 4: Rearrange the equation to gather all terms on one side:
x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64 - 9x - 10 = 0
x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 183x + 54 = 0
Now, we have a sixth-degree equation, and solving it directly can be challenging. So, let's consider the answer choices given.
A. -1, 6
B. [-13 +/- sqrt(193)]/2
C. 1, -6
D. [13 +/- sqrt(145)]/2
To find the correct answer, we can substitute each value into the equation and see which one makes it true. It's a trial and error method, but it can help us get the correct solution.
For example, let's substitute -1 into the equation:
(-1)^6 + 12(-1)^5 + 60(-1)^4 + 160(-1)^3 + 240(-1)^2 + 183(-1) + 54 = 0
1 - 12 + 60 - 160 + 240 - 183 + 54 = 0
The equation doesn't hold true, so -1 is not the solution. We can repeat this process for the other answer choices until we find the solution that satisfies the equation.
It's worth noting that finding the exact solutions for this kind of equation can be quite involved. Therefore, using a numerical or graphical solver may be more efficient if you have access to one.