Solve equation..?

Question

Square root(3x - 2) = 4 - x

Solve for x

Show the work.. I know the answer is 2, but how do you get there??

The answer is two... I need the work to find that answer.

Answer 1

Square root(3x - 2) = 4 - x
Square both side
3x-2 =(4-x)^2
3x - 2 =16 -8x +x^2
0= x^2 -11x +18
0r x^2 -11x +18=0
x^2 -9x -2x+18=0
x(x - 9) -2(x - 9)=0
(x - 2) (x - 9) =0
Therefore x =2 and x=9
But x=9 does not satisfy the equation therefore x=2 is the only answer

Answer 2

square root(3x - 2) = 4 - x

Taking square on both sides we get,

3x -2 = (4 - x) ^2
3x -2 = 16 - 8x - x^2
x^2 - 8x -3x + 16 +2 = 0
x^2 - 11x +18 =0
( x - 9) (x - 2) = 0

x-9 =0 or x-2 =0
x=9 or x=2

The Solution is 2 as it satisfy the given equation.

Answer 3

It's a polynomial equation. therefore in this case a quadratic. so let see:
you can get of the square root by squaring both sides.
you can expand (4-x)^2 to (4^2 - 2*4*x+x^2).
then add and substract
and you will come up with something like this
x^2-11x+18=0
factorizing; (x-2)(x-9)
so values for x are:
x1=2
x2=9

Answer 4

take square of both sides
(3x-2) = (4-x)^2 = 16-8x+x^2
take 3x-2 to the right
0 = 18-11x+x^2
= 18-9x-2x+x^2
= 9(2-x)-x(2-x)
=(9-x) (2-x)
so x = 2 or 9
but x =9 makes the rhs -ve so x=2 is the solution

Note
x^2 = 16 means x= 4 or -4
but sqrt(16) is defined to be positive sqrt so 4

Answer 5

when u take the square root over:

3x - 2 = (4 -x )squared
3x - 2 = (4-x)(4-x)
3x - 2 = 16 -4x -4x -xsquared
3x -2- 16 + 4x +4x + xsquared = 0
xsquared + 11x -18 = 0

then u use the quadratic formula>>>

x= b +/- square root (bsquared - 4ac) over 2ac

where b = 11, c = -18, a =1

x = 1.4462 or -12.446

is it close enough? :p

Answer 6

square root(3x-2) = 4-x
3x-2 = (4-x)(4-x)
3x-2 = 16-8x+x^2
-18 = x^2-8x
x^2-11x+18 = 0
x^2-11x+18 = (x - 2)(x - 9)

Therefore, x = 2 or 9(reject)

Thus, the answer is x = 2.

Answer 7

The answer is not just 2, I believe it's also 9. Square both sides, then re-arrange to be in the form of a classic quadratic equation. Then you'll be able to factor, just like your teachers told you. There is enough information here for you to proceed and do the work yourself. Good luck.

Answer 8

[Square root(3x-2)]Squared=(4-x)squared
3x-2=x Squared - 8x +16
0=x squared -11x +18
0= (x-9)(x-2)
therefore: x=9 & x=2
AND: x=2 is the acceptable value

Answer 9

1st .. square both sides .....
3x - 2 = 16 - 8x + x^2
standard form ... 3x -2 = x^2 - 8x +16
set-up quadratic .... x^2 -11x+18 = 0
factor (x-9) (x-2) = 0 x=(2,9)

check sq rt(3(9)-2) = 4 - 9
sq rt (27-2) = -5
sq rt (25) = plus & minus 5 (check) (-5) (-5) = 25

sqrt (3(2) - 2) = 4 - 2
sqrt (6-2) = 2
sqrt (4) = 2 check

answers are 2 & 9

Answer 10

sqrt(3x - 2) = 4 - x
square both sides
3x - 2 = (4 - x)^2
3x - 2 = (4 - x)(4 - x)
3x - 2 = 16 - 4x - 4x + x^2
3x - 2 = 16 - 8x + x^2
x^2 - 11x + 18 = 0
(x - 9)(x - 2) = 0
x = 2 or 9