# Thread: Couple math problems

1. ## Couple math problems

Need a little help with these two questions:

1. Find x so that the distance between (-3,2) and (x,1) is the square root of 5.

Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:

x^2 + 81 = x^2 + 6x + 9

Not sure what to do there, but I know it's basic algebra.

Any help is appreciated.

2. Originally Posted by amw9490
Need a little help with these two questions:

1. Find x so that the distance between (-3,2) and (x,1) is the square root of 5.

Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:

x^2 + 81 = x^2 + 6x + 9

Not sure what to do there, but I know it's basic algebra.

Any help is appreciated.
Hi amw9490,

I'm sure you know the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute your two points: (-3, 2) and (x, 1)

$\sqrt{5}=\sqrt{(x+3)^2+(1-2)^2}$

$\sqrt{5}=\sqrt{x^2+6x+9+1}$

$\sqrt{5}=\sqrt{x^2+6x+10}$

$5=x^2+6x+10$

$x^2+6x+5=0$

$(x+5)(x+1)=0$

$\boxed{x=-5 \ \ or \ \ x=-1}$

3. Originally Posted by amw9490

Next I had a triangle, needed to find x. Used the pythagorean theorem and got this:

x^2 + 81 = x^2 + 6x + 9

Not sure what to do there, but I know it's basic algebra.

Any help is appreciated.
Assuming what you have done up to this equation is correct, we have

$x^2+81=x^2+6x+9$

If we subtract $x^2$ from both sides we end up with

$81=6x+9$

Subtracting 9 from both sides, we get

$72=6x$

And dividing both sides by 6, we arrive at

$12 = x$