Posted by Brit on Saturday, August 18, 2007 at 10:54pm.
I have a few questions that I need help with.
Solve the radical equation, and check all proposed solutions.
5. x- square root 3x-2=4 I dont understand this problem.
Solve and check the equation.
10. (x^2+14x+49)^3/4-20=7 would I have to multiply each number by 4/3?
Solve the equation by making an apporiate subsistution.
19. x^4 -13x+36=0 I got 2 and 3 for this problem
20. (4x-3)^2-10(4x-3) + 24=0
Solve the absolute value equation or indicate that the equation has no solution.
29. |8x+9| +5 = 9 I got no solution
30. |x^2-4x+4|=2 I got no solution.
5. rewrite it as
x-4 = √(3x-2)
now square both sides
x^2 - 8x + 16 = 3x - 2
x^2 - 11x + 18 = 0
x=9 or x=2
check both answers, x=9 works, x=2 does not.
[(x+7)^2]^(3/4) = 27
(x+7)^(3/2) = 27
[(x+7)^(3/2)]^(2/3) = 27^(2/3)
x+7 = 9
x = 2
19. neither one of your answers satisfy the equation. I was not able to factor it. Check your typing.
I have a feeling it was
x^4 - 13x^2 + 36 = 0
(x^2 - 4)(x^2 - 9) = 0
x=±2 or x=±3
20 let 4x-3 = t
then the equation is
t^2 - 10t + 24 = 0
this factors easily and solves nicely for t.
Once you have the two values of t, sub them back in 4x-3 = t
|8x+9| = 4
8x+9 = 4 or -8x-9 = 4
x = -5/8 or x=-13/8
then x^2-4x+4=2 or -x^2+4x-4 = 2
x^2-4x+2=0 or -x^2+4x-6=0
x^2-4x+2=0 or x^2-4x+6 =0
solve each one using the formula, the first equation yields two irrational answers, the second yields two imaginary answers.
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