square root sign 2x+3 - square root sign x-2 =2
In my head, I worked it as x=3. I did that because only sqr roots that are rational integers can be subtracted to give a whole number.
Now, analytic method. I have no idea, I don't see a simple method in ordinary algebra.
how do you do algebra with integers equations.
To solve algebraic equations with integer coefficients, you can use various methods. Let's go through the steps to solve the equation you provided:
√(2x+3) - √(x-2) = 2
1. Isolate the square root term on one side:
√(2x+3) = √(x-2) + 2
2. Square both sides of the equation to eliminate the square root:
(√(2x+3))^2 = (√(x-2) + 2)^2
3. Simplify both sides of the equation:
2x+3 = (x-2) + 4√(x-2) + 4
4. Gather like terms:
2x+3 = x + 2 + 4√(x-2) + 4
5. Combine like terms on both sides:
2x - x - 2 - 3 - 4 = 4√(x-2)
x - 9 = 4√(x-2)
6. Isolate the square root term:
4√(x-2) = x - 9
7. Square both sides again:
(4√(x-2))^2 = (x - 9)^2
16(x-2) = x^2 - 18x + 81
8. Expand and simplify:
16x - 32 = x^2 - 18x + 81
x^2 - 18x - 16x + 81 + 32 = 0
x^2 - 34x + 113 = 0
9. Use factoring, completing the square, or the quadratic formula to solve for x. In this case, the equation does not factor easily, so the quadratic formula is the best choice:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -34, and c = 113:
x = (34 ± √((-34)^2 - 4(1)(113))) / (2(1))
x = (34 ± √(1156 - 452)) / 2
x = (34 ± √(704)) / 2
x = (34 ± 2√(176)) / 2
x = 17 ± √(176)
So, the solution to the equation √(2x+3) - √(x-2) = 2 is x = 17 + √(176) or x = 17 - √(176).