Solve each equation by using square roots, show all work.
2x^2-5=27
hi again
so first, you need to add 5 to both sides of the equation, so it becomes
2x^2 = 32
next, you divide BOTh sides by 2 because you are going reverse order of operations. this gets rid of the 2 in front of the x^2. so the equation becomes
x^2 = 32/2 which is
x^2 = 16
next, you find the square root of both sides. i dont have a symbol on my keyboard for square root but lets pretend
ok so the square root of x^2 is x. then you must find the square root of 16 (the other side of the equation), but when you take square root, it become PLUS or MINUS . so it would be PLUS or MINUS the square root of 16.
the positive square root of 16 is 4,and the negative of the square root of 16 is -4.
your answers are x = 4 and x = -4 :D
message back on this post if u have more questions
To solve the equation 2x^2 - 5 = 27 by using square roots, we'll follow these steps:
Step 1: Isolate the squared term.
Add 5 to both sides of the equation:
2x^2 - 5 + 5 = 27 + 5
2x^2 = 32
Step 2: Divide both sides by the coefficient of x^2 (2).
Divide both sides of the equation by 2:
2x^2 / 2 = 32 / 2
x^2 = 16
Step 3: Take the square root of both sides.
Take the square root of both sides of the equation:
√(x^2) = √16
x = ±√16
Step 4: Simplify the square root of 16.
The square root of 16 is equal to ± 4, since the square root of any positive number has both positive and negative solutions.
Therefore, the solution to the equation 2x^2 - 5 = 27 by using square roots is:
x = 4 or x = -4.