# 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 don't 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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## 5. To solve the radical equation x - sqrt(3x - 2) = 4, we can rewrite it as x - 4 = sqrt(3x - 2). Now, square both sides of the equation:

(x - 4)^2 = (sqrt(3x - 2))^2

x^2 - 8x + 16 = 3x - 2

x^2 - 11x + 18 = 0

Now we can factorize the quadratic equation by finding two numbers whose product is 18 and sum is -11. By factoring, we get:

(x - 9)(x - 2) = 0

This equation has two solutions: x = 9 and x = 2. To check these solutions, substitute them back into the original equation:

For x = 9: 9 - sqrt(3(9) - 2) = 4 -> 9 - sqrt(27 - 2) = 4 -> 9 - sqrt(25) = 4 -> 9 - 5 = 4 -> 4 = 4 (true)

For x = 2: 2 - sqrt(3(2) - 2) = 4 -> 2 - sqrt(6 - 2) = 4 -> 2 - sqrt(4) = 4 -> 2 - 2 = 4 -> 0 = 4 (false)

Therefore, the solution to the original equation is x = 9.

10. The equation is (x^2 + 14x + 49)^(3/4) - 20 = 7. To solve this equation, we can proceed as follows:

First, simplify the equation by raising both sides to the power of 4/3:

[(x^2 + 14x + 49)^(3/4)]^(4/3) = 27^(4/3)

(x^2 + 14x + 49)^(3/2) = 27^(4/3)

Next, simplify further by taking the square root of both sides:

[(x + 7)^(3/2)]^(2/3) = 27^(2/3)

x + 7 = 9

x = 2

Therefore, the solution to the equation is x = 2.

19. The equation is x^4 - 13x + 36 = 0. The given solutions of x = 2 and x = 3 do not satisfy the equation. It seems like there might be a mistake in the equation provided. Please double-check the equation.

20. Let's make the substitution 4x - 3 = t. The equation becomes t^2 - 10t + 24 = 0. We can now solve this quadratic equation by factoring:

(t - 6)(t - 4) = 0

This equation has two solutions: t = 6 and t = 4. Now, substitute back the original variable:

For t = 6: 4x - 3 = 6 -> 4x = 9 -> x = 9/4

For t = 4: 4x - 3 = 4 -> 4x = 7 -> x = 7/4

Therefore, the solution to the original equation is x = 9/4 or x = 7/4.

29. The equation is |8x + 9| + 5 = 9. To solve this equation, we can proceed as follows:

First, isolate the absolute value term by subtracting 5 from both sides:

|8x + 9| = 4

Now, consider two cases:

Case 1: (8x + 9) is positive:

In this case, we can rewrite the equation as 8x + 9 = 4. Then, solve for x:

8x + 9 = 4

8x = -5

x = -5/8

Case 2: (8x + 9) is negative:

In this case, we can rewrite the equation as -(8x + 9) = 4. Then, solve for x:

-(8x + 9) = 4

8x + 9 = -4

8x = -13

x = -13/8

Therefore, the solutions to the equation are x = -5/8 and x = -13/8.

30. The equation is |x^2 - 4x + 4| = 2. To solve this equation, we can proceed as follows:

First, consider two cases:

Case 1: (x^2 - 4x + 4) is positive:

In this case, we can rewrite the equation as x^2 - 4x + 4 = 2. Then, solve for x:

x^2 - 4x + 4 - 2 = 0

x^2 - 4x + 2 = 0

This equation yields two irrational solutions.

Case 2: -(x^2 - 4x + 4) is positive:

In this case, we can rewrite the equation as -(x^2 - 4x + 4) = 2. Then, solve for x:

-x^2 + 4x - 4 - 2 = 0

-x^2 + 4x - 6 = 0

This equation yields two imaginary solutions.

Therefore, the equation has no real solutions.