Question

if 18 is subtracted from twice the square of an integer, the result is equal to nine times the integer. the integer must be

Answer 1

Okay for this type of problems. let us assume the integer number be x

Answer 2

then square of number becomes x^2
and twice the square of number becomes 2x^2

Answer 3

Let's give this integer a name. Let's call it x.

What is happening to x here? You are subtracting 18 from twice the square of 'x', and that's equal to 9 times x? Can you set up an equation for that?

@shw18

Answer 4

now as he said that if 18 subtracted then, what we would have as equation is
2x^2 -18 right?

Answer 5

Now that is equals to the nine times integer which 9x
therefore
2x^2 -18 = 9x

Answer 6

Solve for x and you will get two posible numbers and they are the solution

Answer 7

2x ^{2}-18=9x ? @genius12

Answer 8

Yes @shw18

Answer 9

et x = the integer:

2x^2 - 18 = 9x
2x^2 - 9x - 18 = 0

Now either factor that or use the quadratic formula if you're lazy like me.

x = -3/2 and 6, so x must be 6 because it has to be an integer.

Done!

Answer 10

after i hav my equation though where do i go from there