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)(x-2)=0

x=9 or x=2

check both answers, x=9 works, x=2 does not.

10.

(x^2+14x+49)^3/4-20=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 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

29.

rewrite as

|8x+9| = 4

8x+9 = 4 or -8x-9 = 4

x = -5/8 or x=-13/8

30.

|x^2-4x+4|=2

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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