Thread: solve radical expression

solve:
sq rt of 5x^2 -7 = 2x
and
sq rt of 1-2x = 1+x

Originally Posted by Belanova

solve:
sq rt of 5x^2 -7 = 2x
and
sq rt of 1-2x = 1+x

do you mean:



and

?

if so, just square both sides and continue. for example, after squaring both sides, the equation becomes . now solve for x. (be sure to check your solutions in the original equation)

So for the first one..
x^2 -7 =0
+7 +7
x=7
no solution

And the second one turns into
1 - 2x = 1 + x^2
-1 = -1
-----------------
-2x = x^2
divide both by 2
-2 = x
no solution

Originally Posted by Belanova

So for the first one..
x^2 -7 =0
+7 +7
x=7
no solution

this is wrong. you have x^2 = 7, not x = 7

And the second one turns into
1 - 2x = 1 + x^2
-1 = -1
-----------------
-2x = x^2
divide both by 2
-2 = x
no solution

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?

the questions are how you typed them. i thought i had to square square 1 and x..
1-2x = 1 + 2x + x^2
..
-4x = x^2 divide by x
-4 = x ; no solution

Originally Posted by Belanova

the questions are how you typed them. i thought i had to square square 1 and x..
1-2x = 1 + 2x + x^2
..
-4x = x^2 divide by x
-4 = x ; no solution

what is the solution for the first?

you cannot divide by x like that.








or

(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 solution for both? :|

1. x^2 =7 cnt divide by x
2. x+4=0
x=-4 = no solution

Originally Posted by Belanova

no solution for both? :|

1. x^2 =7 cnt divide by x
2. x+4=0
x=-4 = no solution

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.

. check both these solutions.

you have and as possible solutions to the second. check both these solutions