Math question...please help :(?

Question

The area of a square with side 2x-1 is 49. Find x.

(2x-1)^2 = 4x^2+1=49 = 4x^2=48

ok now im stuck, then what do you do? btw ^2 means squared.

THANK YOU!

Answer 1

If the square has area 49, it has side sqrt(49) = 7

so 2x - 1 = 7

x  = 4

you could do it the hard way too

(2x - 1)² = 49

4x² - 4x + 1 = 49

4x² - 4x - 48 = 0

x² - x - 12 = 0

(x+3)(x-4) = 0

x = -3, 4

Answer 2

2x-1=7

x=4

Answer 3

woa, slow down the square root of 49 is 7 therefore:

2x-1 = 7

2x = 8

x = 4

also (2x-1)^2 = 4x^2 - 4x + 1   NOT 4x^2 + 1

Answer 4

(2x-1)^2 is actually,

4x^2 - 4x +1

(a-b)^2=a^2 - 2ab + b^2

4x^2 - 4x + 1=49

=4x^2 - 4x - 48=0           (Quadratic Equation)

(4x-16)(x+3)= 0

leading to, x= 4 or -3(which is rejected as it is not possible)

therefore x = 4

Answer 5

If the area = 49, one side must be its root.

Root of 49 = 7.

therefore one side, 2x - 1 = 7

2x = 7+1 = 8

so x = 8/2

       = 4