Given: square root of 5 minus x is equals to x plus 1. Show that the solution to the above equation lies in the interval of negative 1less than or equals to x less than or equals to 5. Solve the equation. Solve the equation negative square root of 5 minus x is equals to x plus one
√(5-x) = x+1
To be defined, 5-x ≥ 0
-x ≥ -5
x ≤ 5
also since √anything yields a positive result, x+1≥0
x≥-1
so -1 ≤ x ≤ 5
To solve, square both sides
5-x = x^2 + 2x + 1
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1
because of our domain, x = 1
To show that the solution lies in the interval of -1 ≤ x ≤ 5, we can first solve the equation step-by-step.
Given: √5 - x = x + 1
Step 1: Let's isolate the terms involving x.
Rearranging the equation, we get:
√5 - 1 = x + x
Step 2: Simplify the left side.
√5 - 1 = 2x
Step 3: Solve for x.
Add 1 to both sides of the equation:
√5 = 2x + 1
Now, square both sides to eliminate the square root:
(√5)^2 = (2x + 1)^2
5 = 4x^2 + 4x + 1
Step 4: Simplify the equation.
Rearrange in standard quadratic form:
4x^2 + 4x + 1 - 5 = 0
4x^2 + 4x - 4 = 0
Step 5: Divide the entire equation by 4 to simplify further:
x^2 + x - 1 = 0
Step 6: Solve the quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation, a = 1, b = 1, and c = -1.
x = (-(1) ± √(1^2 - 4(1)(-1))) / (2(1))
x = (-1 ± √(1 + 4)) / 2
x = (-1 ± √5) / 2
Therefore, there are two solutions:
1) x = (-1 + √5) / 2
2) x = (-1 - √5) / 2
To see if these solutions lie within -1 ≤ x ≤ 5, we can evaluate them:
For x = (-1 + √5) / 2:
(-1 + √5) / 2 ≈ 0.61803398875 (approximately)
For x = (-1 - √5) / 2:
(-1 - √5) / 2 ≈ -1.61803398875 (approximately)
Since both solutions fall within the interval -1 ≤ x ≤ 5, we can conclude that the solution to the equation lies in this specific interval.
To show that the solution to the equation lies in the interval of -1 ≤ x ≤ 5, we need to prove that the values of x within this interval satisfy the equation.
Start by assuming that -1 ≤ x ≤ 5, and substitute x = -1 into the equation:
√5 - (-1) = -1 + 1 --> √5 + 1 = 0
Since the left side of the equation is always positive (√5 is greater than zero), the equation cannot be satisfied when x = -1. Therefore, we can exclude the value -1 from the interval.
Next, substitute x = 5 into the equation:
√5 - 5 = 5 + 1 --> √5 - 5 = 6
Since the left side of the equation is always negative (√5 is less than 5), the equation cannot be satisfied when x = 5. Hence, we can exclude the value 5 from the interval.
So, we have narrowed down the interval to -1 < x < 5. Now let's solve the equation.
Given: √5 - x = x + 1
Rearrange the equation to isolate the square root term:
√5 - 1 = x + x
Combine like terms:
√5 - 1 = 2x
Divide both sides by 2:
(√5 - 1) / 2 = x
Therefore, the solution to the equation is:
x = (√5 - 1) / 2