I have a few questions that I need help with.
Solve the radical equation, and check all proposed solutions.
5. x square root 3x2=4 I don't understand this problem.
Solve and check the equation.
10. (x^2+14x+49)^3/420=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. (4x3)^210(4x3) + 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^24x+4=2 I got no solution.
5. rewrite it as
x4 = √(3x2)
now square both sides
x^2  8x + 16 = 3x  2
x^2  11x + 18 = 0
(x9)(x2)=0
x=9 or x=2
check both answers, x=9 works, x=2 does not.
10.
(x^2+14x+49)^3/420=7
[(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
then
(x^2  4)(x^2  9) = 0
x=±2 or x=±3
20 let 4x3 = 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 4x3 = t
29.
rewrite as
8x+9 = 4
8x+9 = 4 or 8x9 = 4
x = 5/8 or x=13/8
30.
x^24x+4=2
then x^24x+4=2 or x^2+4x4 = 2
x^24x+2=0 or x^2+4x6=0
x^24x+2=0 or x^24x+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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2 answers

khomqi pivtfjr hvfuzt hfjrdgoyt tbdesvi jgepmi ilkn
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answered by fqkiarmon xbku 
shhhhhhkabob
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answered by blob
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