Hi there. My professor gave my class the practice problems for our final exam. I am a little stumped on a few of them and just need someone to help jog my memory. If you have the time, could you show the steps so I can see where I went wrong and practice it. Thank you in advance, I really appreciate it.
1. 4-3(x+1)=2x
For this I got -3/5, but the answer is 1/5
2. 15x^2=x=2
Ans. 1+/-:(radical:-419) over 30 correct is -2/5,1/3
3. (:radical:4x+5)=x
ans. -5/3 correct is 5(-1 is ext)
4-3(x+1)=2x
4-3x-3 = 2x
4-3 = 5x
1 = 5x
x = 1/5
15x^2 + x = 2
15x^2 + x - 2 = 0
x = (-1±√(1+120))/30
= (-1±11)/30
= -12/30,10/30 = -2/5, 1/3
√(4x+5) = x
4x+5 = x^2
x^2-4x-5 = 0
(x-5)(x+1) = 0
x = -1, 5
only x=5 is a solution, since
√(-4+5) = √1 = 1, not -1
of course, squaring makes 1^2 = (-1)^2, but that is not the original equation.
#1
4 - 3(x+1) = 2x
4 - 3x - 3 = 2x
-5x = -1
x = -1/-5 = 1/5
#2
15x^2=x=2 , assuming you meant 15x^2 + x=2
since the = and + sign are on the same key
15x^2 + x - 2 = 0
which factors to (3x-1)(5x+2) = 0
x = 1/3 or x = -2/5
by formula: x = (-1 ± √121)/30 = 1/3 or -2/5
√(4x+5) = x
square both sides
4x+5 = x^2
x^2 - 4x - 5 = 0
(x-5)(x+1) = 0
x = 5 or x = -1
but since you squared ALL answers must be verified
if x = 5
LS = √25 = 5
RS = 5 , so that's good
if x = -1
LS = √4 = 2
RS =1 ≠LS
so x = 5
Of course! I'd be happy to help you with these practice problems and walk you through the steps. Let's go through each problem together:
1. 4-3(x+1) = 2x
To solve this equation, we need to simplify and isolate the variable x. Let's start by distributing the -3 to both terms inside the parentheses:
4 - 3x - 3 = 2x
Next, combine like terms:
1 - 3x = 2x
To isolate x, we can move all the x terms to one side by adding 3x to both sides:
1 - 3x + 3x = 2x + 3x
1 = 5x
Now, divide both sides by 5 to solve for x:
1/5 = x
So, the correct answer is x = 1/5. It seems that you made an error somewhere in your calculation.
Moving on to the next problem:
2. 15x^2 = x + 2
This equation is a quadratic equation in the form of ax^2 + bx + c = 0. To solve it, we need to rearrange the equation to bring all the terms to one side and then factor or use the quadratic formula.
15x^2 - x - 2 = 0
Now, we can either factor this equation or use the quadratic formula. Let's use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
Comparing this to our equation, we have:
a = 15, b = -1, c = -2
Substituting these values into the quadratic formula:
x = [-(-1) ± sqrt((-1)^2 - 4(15)(-2))] / (2(15))
Simplifying:
x = [1 ± sqrt(1 + 120)] / 30
x = [1 ± sqrt(121)] / 30
x = [1 ± 11] / 30
So, the two solutions are:
x = (1 + 11) / 30 = 12/30 = 2/5
x = (1 - 11) / 30 = -10/30 = -1/3
Therefore, the correct answers are x = -2/5 and x = 1/3.
Now let's move on to the last problem:
3. sqrt(4x + 5) = x
To solve this equation, we need to isolate the square root term and then square both sides to eliminate the square root. Let's go step by step:
Step 1: Isolate the square root term
Start by subtracting x from both sides:
sqrt(4x + 5) - x = 0
Step 2: Square both sides
(square both sides to eliminate the square root)
(sqrt(4x + 5) - x)^2 = 0^2
(4x + 5) - 2x(sqrt(4x + 5)) + x^2 = 0
Step 3: Simplify the equation
Combine like terms:
x^2 - x(sqrt(4x + 5)) + (4x + 5) = 0
Step 4: Factor the equation if possible
However, this equation cannot be easily factored. So, we can use the quadratic formula to solve for x similar to problem 2.
x = [-b ± sqrt(b^2 - 4ac)] / 2a
Comparing this to our equation, we have:
a = 1, b = -sqrt(4x+5), c = 4x+5
x = [sqrt(4x+5) ± sqrt((sqrt(4x+5))^2 - 4(1)(4x+5))]/(2(1))
Simplifying:
x = [sqrt(4x+5) ± sqrt(4x+5 - 16x - 20)]/2
x = [sqrt(4x+5) ± sqrt(-12x - 15)]/2
Notice that we have the square root of a negative number, which means that the equation has no real solutions. However, it does have complex solutions. It seems that you mistakenly wrote (-5/3) instead of 5 where you missed a negative sign. The correct answer is x = 5.
I hope this explanation helps clarify the steps and answers for these practice problems. If you have any further questions or need more assistance, please let me know!