please solve this 3x² - 1 =47?

Question

the answer should be one of these:
a) {-4,4}
b) {-16,16}
c) {-256, 256}
d) {0,4}

please show the solution

Answer 1

first ans is correct,,,,,,,,,,

oh but so many other people already answered the ques in a min. so i dont wanna solve it for u again.

But felt gud solving it in my mind after 19 years of leaving college and about 22 years of leaving HSC.

Answer 2

3x² - 1 =47
3x² = 47 + 1
3x² = 48 /:3
x² = 16 /sqrt
x1,2 = +/-4
x1 = 4
x2 = -4

check:
3*4^2 -1 = 47
3*16 - 1 = 47
48 -1 = 47
47 = 47 => works for x = 4, and it will work for -4 as well, since sq(-4) = sq (4)

therefore your solution is a) {-4,4}

Answer 3

3x² - 1 =47 or
3x^2=48 or x^2=16 which leads to :
x= +4 or - 4.
Answer: letter (a).

Answer 4

Subtract 47 from both sides:
3x^2-48=0

Factor out a 3:
3(x^2-16)
(x+4)(x-4)=0
x=-4 and 4

The answer is a.

I hope this helps!

Answer 5

3x² - 1 = 47

3x² - 1 + 1 = 47 + 1

3x² = 48

3x² / 3 = 48 / 3

x² = 16

√x² = √16

x = ± 4

- - - - - - -

Check for -

3x² - 1 = 47

3(- 4)² - 1 = 47

3(16) - 1 = 47

48 - 1 = 47

47 = 47

- - - - - - - -

Check for +

3x² - 1 = 47

3(4)² - 1 = 47

3(16) - 1 = 47

48 - 1 = 47

47 = 47

- - - - - - - - - - -

The answer is a { - 4, 4 }

- - - - - - -s-

Answer 6

First, add 1 to both sides of the equation:
3x^2 = 48
Now divide each side by 3
x^2 = 16
Now take square roots
x = plus or minus 4
Answer is a

Answer 7

3x^2 - 1 =47
3x^2 = 47+1
3x^2 = 48
x^2 = 48/3
x^2 = 16
x = 4, -4

Answer 8

X1=-4
x2=4

Answer 9

3x^2 -1 =47
3x^2 = 47 + 1
3x^2 = 48 *divide both sides by 3
x^2 = 16
x = +/- 4

Answer 10

3x^2 -1=47
+1 +1
3x^2 =48


3x^2 =48
----- -----
3 3

x^2=16


sqrt(x^2) = sqrt(16)

x= 4 or -4
answer a

Answer 11

3x^2 - 1 = 47

Add 1 to both sides:

3x^2 = 48

Divide both sides by 3:

x^2 = 16

x = + or - sq.root of 16

x = + 4 and - 4

Ans: {-4, 4}