I have never had algebra and cannot figure out how to work through the problem below
[(x + 1)² - (2x + 1)]1/2 + 2|x| - 6 = 0
Can you show me the steps for solving the problem?
I'm not sure whether the 1/2 is supposed to be multiplied or you forgot a sign before it.
IxI = absolute value of x
Get rid of parentheses first.
[(x + 1)² - (2x + 1)] = x^2 +2x +1 - 2x - 1 = x^2
(x+5)(x+2)=X2+7x+10
Of course! I'd be happy to help you solve this algebraic equation step by step.
Step 1: Simplify the Equation
Start by simplifying the equation and removing any unnecessary parentheses or brackets. Let's break it down:
[(x + 1)² - (2x + 1)]1/2 + 2|x| - 6 = 0
First, simplify the terms within the square brackets:
(x + 1)² = (x + 1)(x + 1) = x² + 2x + 1
(2x + 1) = 2x + 1
Now, substitute the simplified terms back into the equation:
[(x² + 2x + 1) - (2x + 1)]1/2 + 2|x| - 6 = 0
Simplifying further:
(x² + 2x + 1) - (2x + 1) = x² + 2x + 1 - 2x - 1 = x²
The equation is now simplified to:
x²1/2 + 2|x| - 6 = 0
Step 2: Isolate the Absolute Value
The equation now contains an absolute value term (2|x|). To solve for x, we need to isolate the absolute value term. To do this, we'll consider two cases: one where |x| is positive and one where |x| is negative.
Case 1: |x| = x
In this case, rewrite the equation without the absolute value symbols:
x²1/2 + 2x - 6 = 0
Case 2: |x| = -x
In this case, rewrite the equation by replacing |x| with -x:
x²1/2 + 2(-x) - 6 = 0
x²1/2 - 2x - 6 = 0
Step 3: Solve for x in Case 1
Let's solve the first case:
x²1/2 + 2x - 6 = 0
This is a quadratic equation in the form of ax² + bx + c = 0, where a = 1/2, b = 2, and c = -6.
To solve the equation, you have a few options: factoring, completing the square, or using the quadratic formula. Since factoring this quadratic equation may not be straightforward, we'll use the quadratic formula:
x = (-b ± √(b² - 4ac))/(2a)
Plugging in the values from our equation, we have:
x = (-(2) ± √((2)² - 4(1/2)(-6)))/(2(1/2))
x = (-2 ± √(4 + 12))/1
x = (-2 ± √(16))/1
x = (-2 ± 4)/1
Now, evaluate the positive and negative solutions separately:
x₁ = (-2 + 4)/1 = 2/3
x₂ = (-2 - 4)/1 = -6
So, in the first case, we have two solutions: x₁ = 2/3 and x₂ = -6.
Step 4: Solve for x in Case 2
Now let's solve the second case:
x²1/2 - 2x - 6 = 0
Again, this is a quadratic equation in the form of ax² + bx + c = 0, where a = 1/2, b = -2, and c = -6.
Using the quadratic formula, we have:
x = (-b ± √(b² - 4ac))/(2a)
Plugging in the values, we get:
x = (-(-2) ± √((-2)² - 4(1/2)(-6)))/(2(1/2))
x = (2 ± √(4 + 12))/1
x = (2 ± √(16))/1
x = (2 ± 4)/1
Again, evaluate the positive and negative solutions separately:
x₃ = (2 + 4)/1 = 6
x₄ = (2 - 4)/1 = -2
So, in the second case, we have two more solutions: x₃ = 6 and x₄ = -2.
Step 5: Final Solution
Combining the solutions from both cases, we have a total of four solutions to the original equation:
x₁ = 2/3
x₂ = -6
x₃ = 6
x₄ = -2
These are the values of x that satisfy the given equation [(x + 1)² - (2x + 1)]1/2 + 2|x| - 6 = 0.