# Thanks for your help in advance. The area of a square with side 2x-1 is 49. Find x.

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### 2 Answers

To solve for the value of x, apply the formula of area of square.

`A= s^2`

where s represents the length of the side of the square.

So, plug-in A=49 and s=2x-1 to the formula.

`49=(2x-1)^2`

Then, expand right side of the equation.

`49=4x^2-4x+1`

Set one side of the equation equal to to zero. To do so, subtract both sides by 49.

`49-49=4x^2-4x+1-49`

`0=4x^2-4x-48`

To simplify, divide both sides by the GCF of 4x^2-4x-48.

`0=4(x^2-x-12)`

`0/4=(4(x^2-x-12))/4`

`0=x^2-x-12`

Then, factor.

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

Then, set each factor equal to zero and solve for x.

For the first factor:

`x-4=0`

`x-4+4=0+4`

`x=4`

And for the second factor:

`x+3=0`

`x+3-3=0-3`

`x=-3`

To determine which of these two values of x should be considered as the solution, plug-in the values to 2x-1. But, take into consideration that 2x-1 represents the length of the side of the square. So, the resulting value should be positive (greater than 0).

For the first value x=4:

`2x - 1gt0`

`2(4) - 1gt0`

`7gt 0` (True)

For the second value x=-3:

`2x - 1gt0`

`2(-3)-1gt0`

`-7gt0` (False)

**Hence, the value of x in a square that has a side 2x-1 and an area of 49 is 4 `(x=4)` . **

The area of a square can be found by `A=s^2` where A is the area and s the side length.

Here A=49 and s=2x-1 so:

`49=(2x-1)^2`

`49=4x^2-4x+1`

`4x^2-4x-48=0`

`4(x^2-x-12)=0`

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

So x=4 or -3. Since the value of 2x-1 has to be positive (all lengths are nonnegative) we take x=4.

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The solution is x=4

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Check: 2(4)-1=7 and 7*7=49

It might be easier to do this problem if you realize that if the area is 49, the sides must be 7. Then 2x-1=7 ==> x=4