Question

HELP ME PLEASE <3 find to the nearest tenth, the positive value of x in the equation sqrt x^2+21 =2x
1. 7
2. 4.6
3. 2.6
4. 1.3

Answer 1

Is 21 in the square root or not?

Answer 2

x=21 I removed the square root and got that for x

Answer 3

its in square root :)

Answer 4

Ok. Then square both side first. You get x^2 + 21 = 4x^2.

Simplify. 3x^2 = 21
Divide each side by 3. x^2 = 7. Your answer is 2.6 approximately.

Answer 5

another way:

square both sides to get x^2+21 = 4x^2 . after that ,combine like terms 3x^2 = 21
x^2 = 7
after you solve for x, you get x = sqrt(7) . also known as 2.6 :)

Answer 6

Thank you guys!! xo

Answer 7

No problem! :)