## Tutor Answer

I am assuming that 'x' and 'a' are the variables we are using for the solutions.

The equation representing the difference between both solutions would probably be:

a - x = 30

Since we are given that x^2 = a, that means that a = x^2. And we can substitute into the equation and then solve for x.

*a - x = 30 <Difference Equation>*

* x^2 - x = 30* *<Since a = x^2, we can replace a with x^2>*

* Equation is now quadratic, so we can solve for x*

* x^2 - x - 30 = 30 - 30 <Subtract both sides by 30>*

* x^2 - x - 30 = 0*

* (x + 5)(x - 6) = 0 <Left side is factored>*

* x + 5 = 0 or x - 6 = 0* *<Zero Product Property>*

* x + 5 - 5 = 0 - 5 or x - 6 + 6 = 0 + 6 <Solve each equation separately>*

* x = -5 or x = 6*

* I am thinking we need to use the positive x-value, so we can try x = 6*

Now that we have our value of x, we can substitute back into our difference equation to solve for 'a'

*a - x = 30 <Difference Equation>*

* a - 6 = 30 <Substituting our calculated value of x, which is x = 6>*

* a - 6 + 6 = 30 + 6 <Add both sides by 6>*

* a = 36*

**POSSIBLE SOLUTION: a = 36 (Only if you are required to use just the positive x value)**

I don't know if your problem requires more than one answer. But just in case, we will also test x = -5

*a - x = 30 <Difference Equation>*

* a - (-5) = 30 <Substitute x = -5>*

* a + 5 = 30 < A - (-B) is the same as A + B>*

* a + 5 - 5 = 30 - 5 <Subtract both sides by 5>*

* a = 25*

**POSSIBLE SOLUTION: a = 36 or a = 25 (Only if you needed more than one solution)**

I hope this helps.

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