i'm right! :)
- Thread starter humakhan
- Start date
Sorry but...
3x = 18 - x^2
take x^2 to other side and subtract
so then its.
x^2+3x = 18 then
x^2+3x-18=0
Can you factor that to get two values for x? You have a head start 'cause you proved that one of them is 3 so one factor is x-3.
but what happened to the 3x???
3x = 18 - x^2
take x^2 to other side and subtract
so then its.
x^2+3x = 18 then
x^2+3x-18=0
Can you factor that to get two values for x? You have a head start 'cause you proved that one of them is 3 so one factor is x-3.
G'day, humakhan!
Unfortunately, you're not right
.
The standard form of a quadratic equation is ax^2 + bx + c = 0. You want to arrange it in this way.
If we go the way you were...
Transfer the x^2 term to the left-hand side by adding it to both sides:
3x + x^2 = 18
Transfer 18 to the left-hand side by subtracting:
3x + x^2 - 18 = 0
ie. x^2 + 3x -18 = 0
Then can you factorise and solve for x?
We cannot subtract 3x from x^2 (or whatever it was you did :wink because x is not necessarily the same as x^2 just as 3^2=9 is not equal to 3.
Edit: Gene <-> too quick.
Edit: Can't get much closer! Gene
Unfortunately, you're not right
.(-3)^2 also equals 9. The answer is x=3 or x=-3. If you check x=-3 as you did with x=3, you'll see that it doesn't work for the original equation..
You need to recognise this as a quadratic equation. We have an x^2 term.
The standard form of a quadratic equation is ax^2 + bx + c = 0. You want to arrange it in this way.
If we go the way you were...
Transfer the x^2 term to the left-hand side by adding it to both sides:
3x + x^2 = 18
Transfer 18 to the left-hand side by subtracting:
3x + x^2 - 18 = 0
ie. x^2 + 3x -18 = 0
Then can you factorise and solve for x?
We cannot subtract 3x from x^2 (or whatever it was you did :wink
because x is not necessarily the same as x^2 just as 3^2=9 is not equal to 3.Edit: Gene <-> too quick.
Edit: Can't get much closer! Gene