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March 26th, 2014, 11:48 PM   #1
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Clearing fractions in a quadratic equation

The equation is y = - x^2 / 40 + 31x / 40 + 4/5

I clear the fractions by multiplying through by 40 and obtain a solution of x = -1 and x = 32 for the x intercepts. However when I multiply the equation through by 40 the y intercept becomes 32 and is wrong. I thought I had to multiply both sides by 40 which also means I end up with 40y and a y intercept 40 times larger than it should be. Any help would be appreciated.
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March 27th, 2014, 12:12 AM   #2
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Re: Clearing fractions in a quadratic equation

How are you calculating the y-intercept?
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March 27th, 2014, 12:14 AM   #3
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Re: Clearing fractions in a quadratic equation

by clearing the fractions, so multiplying the whole equation by 40, giving -x^2 + 31x + 32
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March 27th, 2014, 12:15 AM   #4
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Re: Clearing fractions in a quadratic equation

Then substituting x = 0 into said equation
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March 27th, 2014, 12:27 AM   #5
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Re: Clearing fractions in a quadratic equation

Quote:
Originally Posted by welshwizard1981
by clearing the fractions, so multiplying the whole equation by 40, giving -x^2 + 31x + 32
That's not an equation! That's just an expression in x. Where's y?
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March 27th, 2014, 12:32 AM   #6
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Re: Clearing fractions in a quadratic equation

The question is y = -x^2 / 40 + 31x / 40 + 4/5
I multiply the equation through by 40 which gives:

40(y) = -x^2 + 31x + 32
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March 27th, 2014, 12:35 AM   #7
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Re: Clearing fractions in a quadratic equation

If y = 0 then 40 x y = 0 so it is in quadratic format
0 = -x^2 + 31x + 32

When I use a graph app, the solution of x is correct x = -1 and x = 32 but the y intercept is wrong, it is not 32 but remains as 4/5 but I do not understand how when your supposed to multiply all of the equation by 40
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March 27th, 2014, 12:51 AM   #8
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Re: Clearing fractions in a quadratic equation

If you multiply an equation by 40, you multiply all of an equation by 40. You can't multiply part of an equation by 40 and not the rest of it. So, you multiplied by 40 and got:



To find the y-intercept, you set x = 0, which gives:



Which is exactly the same as putting x = 0 in the original equation.

I suspect you're just typing the x-expression into your computer, which is assuming that's the expression for y. But, of course, that's the expression for 40y.

The problem is perhaps more understanding the software you're using than the maths!
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March 27th, 2014, 01:05 AM   #9
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Re: Clearing fractions in a quadratic equation

Of course, now you have mentioned that i can see it now. I just haven't seen a quadratic with 40y, only with one y. of course its just a case of rearranging the equation. Thanks for your help, appreciate that.
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