Slove using the square root property

[latresa31s]

I have to solve this problem using the square root property; expressing any complex numbers using i notation

(6x + 5)^2 = 2

help please

 

[Unco]

Do you recall what the square root property is?

(Also, there are no non-real roots to have to worry about here.)

 

[stapel]

Take the square root of each side:

. . . . .(6x + 5)² = 2

. . . . .sqrt[(6x + 5)²] = ± sqrt[2]

. . . . .6x + 5 = ± sqrt[2]

...and solve for x.

Hope that helps a bit.

Eliz.

 

[latresa31s]

(6x + 5)^2 = 2
6x + 5 = +-^2
6x = -5+-^2
x = -5+-^2/5

is this right?

 

[stapel]

As has been explained a few times in another thread, squaring something is very different from taking the square root of something. Since these operations are very different, they have different notation.

To indicate a square root, please use the suggested "sqrt()" notation. Please reserve the carat character for its customary meaning of "to the power of".

Since -5 ÷ 6 does not equal -5, and since "± to the power two" (or is it "two-fifths?) has no meaning, it is unlikely that "-5 ± to the power 0.4" would be correct. Sorry.

Please reply using standard notation and grouping symbols. Thank you.

Eliz.

 

[Denis]

(6x + 5)^2 = 2
6x + 5 = +-^2
6x = -5+-^2
x = -5+-^2/5
is this right?

Are you doing this on purpose?
WHY are you using ^2 to represent sqrt(2)?
stapel just told you: 6x + 5 = +- sqrt[2]

using ^2 here is WRONG.

6x + 5 = +- sqrt(2)
6x = -5 +- sqrt(2)
x = [-5 +- sqrt(2)] / 6

Where did the 6 go in your solution? Don't see it anywhere...

 

[latresa31s]

Ok sorry that this has been stress to you guys. I can't seem to understand even how to work the problem . Thanks for all your help :?