solve:
sq rt of 5x^2 -7 = 2x
and
sq rt of 1-2x = 1+x
do you mean:
$\displaystyle \sqrt{5x^2 - 7} = 2x$
and
$\displaystyle \sqrt{1 - 2x} = 1 + x$?
if so, just square both sides and continue. for example, after squaring both sides, the equation becomes $\displaystyle 5x^2 - 7 = 4x^2$. now solve for x. (be sure to check your solutions in the original equation)
this is wrong. you have x^2 = 7, not x = 7
you didn't write x^2 on the right the first time. make sure you are doing the right thing. should it be x^2 or x?And the second one turns into
1 - 2x = 1 + x^2
-1 = -1
-----------------
-2x = x^2
divide both by 2
-2 = x
no solution
what is the solution for the first?
you cannot divide by x like that.
$\displaystyle 1 - 2x = 1 + 2x + x^2$
$\displaystyle \Rightarrow x^2 + 4x = 0$
$\displaystyle \Rightarrow x(x + 4) = 0$
$\displaystyle \Rightarrow x = 0$ or $\displaystyle x = -4$
(since x = 0, it was invalid for you to divide by it. you need to factor by pulling out the common term).
now continue
no. if you have x^2 on one side of an equation, you can square root both sides, as long as the other side is non-negative.
$\displaystyle x^2 = 7 \implies x = \pm \sqrt{7}$. check both these solutions.
you have $\displaystyle x = 0$ and $\displaystyle x = -4$ as possible solutions to the second. check both these solutions