Solve for x is this right?

Question

(2/3)(5x-2)^2+3=4 u can start but multiplying the 3 right so no fractions and you only need to multiply 2/3 x 3 and 3(3) and 4(3) not (5x-2)^2 rite? or u can do this?...(2/3)(5x-2)^2+3=4....

(2/3)(5x-2)^2=1 now multiply by 3 to get rid of denominator 2(5x-2)^2=3 final answer x=sqrt(3/2)/(5)+2/(5) is this right?

Answer 1

Your first assumption is correct...the (5x-2)^2 term is already multiplied by 3 when you multiply 2/3 by 3 (since it is all multiplication).

Your final answer is technically, mostly correct. When you take the square root of something, you get two answers (for instance, sqrt(4) = +/-2, since 2x2 = 4 *and* -2x-2 = +4).

Correction for that:

x= +/- sqrt(3/2)/(5)+2/(5)

Final note...and this is a very "picky" final note: usually fractions are not left under the square root, so they are usually rationalized.

Nice job!

Answer 2

Yes but x has here two solutions.

when you have (5x-2)^2=3/2, then 5x-2 = +/-sqrt(3/2).

the second solution is x = -sqrt(3/2)/5+2/5.