Homework due in 1 hour, please help me solve these problems....
sqrt (x-5) + 14 = 3x + 2
4x - x/3 + (x + 1)/9 = 2x
Isolate the square root:
$\displaystyle \sqrt{x - 5} = 3x - 12$
Now square both sides.
$\displaystyle x - 5 = (3x - 12)^2$
and solve for x normally from there.
Always always always plug your solutions back into the original equation. This method occasionally gives extra solutions that don't work. If none of the solutions work, then there is no solution.
-Dan
Remember you have a square root on that x - 5. It does not cover the 14! See my first response for what I think is the best approach. If you do it your way you do not get rid of the square root:
$\displaystyle \left ( \sqrt{x - 5} + 14 \right ) ^2 = (x - 5) + 2 \cdot 14 \cdot \sqrt{x - 5} + 14^2$
So continuing from where I left off in my first post:
$\displaystyle x - 5 = (3x - 12)^2$
$\displaystyle x - 5 = 9x^2 - 72x + 144$
$\displaystyle 9x^2 - 73x + 149 = 0$
It might factor, it might not. For something with coefficients that look like this I'd just skip right to the quadratic formula to finish it.
(Hint: Check the discriminant. What does that tell you about the possible solutions for x?)
-Dan
First, after rearranging, the equation equals $\displaystyle \sqrt{x-5}=3x-12.$ We require $\displaystyle x-5\ge0$ & $\displaystyle 3x-12\ge0,$ from here, if the solution does exist, it must lie in the interval $\displaystyle [5,\infty[.$ Now note that the equation equals $\displaystyle \sqrt{x-5}-3(x-5)=3,$ and for $\displaystyle x\ge5\implies\sqrt{x-5}\le3(x-5),$ hence the LHS is negative, so the equation has no solution.