Algebra. x^2 = 32, my back of book says +/- 4√2 . But I keep getting x = +/- 2√8.?

Question

What am I doing wrong and why does it still work in the equation? Or can both answers be correct?

I'm very confused, thanks for any help.

Answer 1

Solve for "x" in the equation....

x^2 = 32

First: eliminate the exponent (2) by finding the square root of both sides.

V`(x^2) = +/- V`(32)

Sec: express 32 in lowes terms.

V`(x*x) = +/- V`(4*8)

V`(x*x) = +/- V`(2*2*8)

V`(x*x) = +/- V`(2*2*2*4)

V`(x*x) = +/- V`(2*2*2*2*2)

x = +/- V`(2*2*2*2*2)

Third: each pair of 2's becomes one & you bring it to the front of the radical sign (represented by V`).

x = +/- 2(2) V`(2)

x = +/- 2 V`(2)

Or, x = 2 V`(2) & -2 V`(2)

Answer 2

If x^2 = 32 then x = +- sqrt (32)

sqrt 32 = sqrt(16) times sqrt (2) which is 4 sqrt 2

However, 2 sqrt 8 equals 2 [sqrt (4) times sqrt (2)] which is 2 times 2 times sqrt 2, also 4 sqrt 2

So both are correct but the books answer is simplified completely while yours is only partially simplified

Answer 3

These answers are equivalent. Yours can be simplified a little more...

x^2 = 32
x = ± √(16 * 2)
x = ± 4√2

or

x2 = 32
x = ± √(4 * 8)
x = ± 2√(8)
x = ± 2√(4 * 2)
x = ± 4√(2)

Answer 4

x^2=32 Look for a factor with an even sq rt
√32=√16x√2 √16 = 4

bks answer is correct +- 4√2
Okay?

Answer 5

These answers are identical. The square root of 8 is the same as the square root of (2*4), but the square root of 4 equals 2, so the square root of 8 can be written as 2*square root of 2.

Answer 6

The two answers are equivalent and both are right.
2√8 = 2√(4*2).
Now take the square, 4, out of the radical to get
2*2√2 or 4√2.

Answer 7

you see math is type of subject that always has different forms of calculation and answers, which work out to be right.
For your question you must first move the square to the other side. that gives us √32. now the factors of this number is √8√4. we can now factor them more to give us 4√2. so book is right in this case.
note----- you cannot factor a factor of a radical,

Answer 8

You'd solve this is as the sqrt of 32.

the sqrt of 32 can be written as sqrt(16*2)

16 is a perfect square, so it can be pulled out in front of the sqrt as a 4, leaving the radial 2 behind.

giving you as your answer

+/-4sqrt(2)

Answer 9

The answer in the book is more simplified. Factor out 8 and keep going.

Answer 10

both answers are correct.

+/- 4 2^(1/2) is same as +/- 2(8)^(1/2)

see, u can write 8 = 4 *2

& 4^(1/2)= 2

so (8)^(1/2) = (4*2)^(1/2) =2 (2)^1/2

+/- 2(8)^(1/2) = +/- (2* [2(2)^(1/2)]) = +/- 4(2^(1/2))

so ur answer is correct.