VARIATION: Find all real solutions to the equation: √(variable_93 + 3 - 4√(variable_93 - 1)) + √(variable_93 + 8 - 6√(variable_93 - 1)) = 1. Find the numerical value if exact
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Step-by-Step Solution
Step 1: Simplify the terms under the square roots
We observe that the expressions under the square roots resemble the expansion of a perfect square (a−b)2=a2−2ab+b2. Let a=x−1. Then for the first term, we have x+3−4x−1=(x−1)−4x−1+4=(x−1)2−2(2)(x−1)+22=(x−1−2)2. Similarly, for the second term, x+8−6x−1=(x−1)−6x−1+9=(x−1)2−2(3)(x−1)+32=(x−1−3)2.
Step 2: Apply the property a2=∣a∣
Using the property that a2=∣a∣, we can simplify the equation to involve absolute values. This gives us the sum of two absolute value expressions equal to 1.
Step 3: Define a substitution and analyze cases
To simplify the problem, let's substitute y=x−1. Since x−1 must be non-negative for x−1 to be real, we have x≥1, which implies y≥0. We need to consider different cases for y based on the critical points y=2 and y=3.
Step 4: Solve for y in different cases
We analyze three cases for y: \\ 1. If y<2, then y−2 is negative and y−3 is negative. The equation becomes −(y−2)−(y−3)=1, which simplifies to −2y+5=1, giving y=2. This contradicts our assumption y<2. \\ 2. If 2≤y≤3, then y−2 is non-negative and y−3 is non-positive. The equation becomes (y−2)−(y−3)=1, which simplifies to 1=1. This is always true, so all values of y in the interval [2,3] are solutions. \\ 3. If y>3, then y−2 is positive and y−3 is positive. The equation becomes (y−2)+(y−3)=1, which simplifies to 2y−5=1, giving y=3. This contradicts our assumption y>3. \\ Therefore, the solution for y is 2≤y≤3.
Step 5: Substitute back to find x
Now we substitute back y=x−1 into the inequality 2≤y≤3.
Step 6: Solve for x
To solve for x, we square all parts of the inequality. This gives 4≤x−1≤9. Then, we add 1 to all parts of the inequality to isolate x, resulting in 5≤x≤10. This is the range of all real solutions for x.