find real solutions
sq.root 2x+5 - sq.root x+6 = 1
(2x+5)^.5 - (x+6)^.5 = 1 you mean maybe, parentheses missing if so
(2x+5)^.5 = 1 +(x+6)^.5
2x+5 = 1 + 2 (x+6)^.5 + x+6
(x - 2) = 2 (x+6)^.5
x^2 - 4 x + 4 = 4 x + 24
x^2 - 8 x - 20 = 0
(x-10)(x+2) = 0
x = 10 or x = -2
go back and see what works in the original now.
yes thank you!
To find the real solutions to the equation √(2x + 5) - √(x + 6) = 1, we need to isolate x on one side of the equation. Here's how you can do that:
Step 1: Start by isolating one of the square roots. Let's isolate √(2x + 5):
√(2x + 5) = 1 + √(x + 6)
Step 2: Square both sides of the equation to eliminate the square root:
(√(2x + 5))^2 = (1 + √(x + 6))^2
2x + 5 = 1 + 2√(x + 6) + (x + 6) [Using (a + b)^2 = a^2 + 2ab + b^2]
Step 3: Simplify the equation:
2x + 5 = 1 + 2√(x + 6) + x + 6
2x + 5 = 7 + 2√(x + 6) + x
Step 4: Rearrange the equation:
2x - x + 5 - 7 - √(x + 6) = 2√(x + 6)
Step 5: Simplify further:
x - 2 - √(x + 6) = 2√(x + 6)
x - √(x + 6) = 2√(x + 6) + 2
Step 6: Bring all terms with square root on one side:
x - 2√(x + 6) - √(x + 6) = 2
Step 7: Simplify and combine like terms:
x - 3√(x + 6) = 2
Step 8: Square both sides to eliminate the square root:
(x - 3√(x + 6))^2 = 2^2
x^2 - 6x√(x + 6) + 9(x + 6) = 4
Step 9: Expand and simplify the left side of the equation:
x^2 - 6x√(x + 6) + 9x + 54 = 4
x^2 + 9x - 4 - 6x√(x + 6) - 54 = 0
x^2 + 9x - 58 - 6x√(x + 6) = 0
Step 10: Now, we have a quadratic equation in terms of x. To solve this equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
For this equation, a = 1, b = 9, and c = -58. Substituting these values into the quadratic formula, we get:
x = (-9 ± √(9^2 - 4(1)(-58))) / (2(1))
Step 11: Simplify the equation:
x = (-9 ± √(81 + 232)) / 2
x = (-9 ± √313) / 2
Therefore, the real solutions for the equation are:
x = (-9 + √313) / 2
x = (-9 - √313) / 2
These are the real solutions for the given equation.