Grouping like terms together
Factoring
Now square both sides
I'm taking Pre-Calc and Trigonometry, and the class is so poorly put together I don't even get a textbook. My teacher is of no help. I can't figure out how to solve this problem, and I desperately would like to understand. Help?
√x+2 + √3x+4 = 2
I have tried it over and over and have gotten different answers each time. I don't know what are the proper steps to solve an equation with two radicals on one side. :<
Or this might be the right form? √(x+2) + √(3x+4) = 2.
squaring BS,
(x + 2) + 2√(x + 2)√(3x + 4) + 3x + 4 = 4, simplify
2√(x + 2)√(3x + 4) = 4 - 4 - 2 - x - 3x,
2√(x + 2)√(3x + 4) = -2 -4x = -2(1 + 2x)
√(x + 2)√(3x + 4) = -(1 + 2x), square BS,
(x + 2)(3x + 4) = (-1 - 2x)^2,
3x^2 + 10 x + 8 = 1 + 4x + 4x^2, combine like terms
(4x^2 - 3x^2) + (-10x + 4x) + (1 - 8) = 0,
x^2 - 6x - 7 = 0, factoring
(x - 7)(x + 1) = 0, using the zero factor theorem
x = -1 or x = 7.
Looking for extraneous root, we substitute the value to the original equation.
(x + 2)^(1/2) + (3x + 4)^(1/2) = 2, for x = 7
(7 + 2)^(1/2) + (3(7) + 4)^(1/2) = 2?
3 + 5 = 2? not the root of the original equation,
Next, try x = -1.
(- + 2)^(1/2) + (3(-1) + 4)^(1/2) = 2?
1 + 1 = 2. x = -1 is the ONLY ROOT of the above equation.
Check by graphing: surely x = -1 is a ROOT.