Not sure how to work this
- Thread starter Richard B
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D
Deleted member 4993
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Please follow the rules of posting in this forum, as enunciated at:Find all values ofsuch that
.![\[\frac{x}{x - 5} = \frac{4}{x - 4}.\]](https://latex.artofproblemsolving.com/7/b/e/7be29781055021baf4416f35f07c31d19fc7c08d.png "\[\frac{x}{x - 5} = \frac{4}{x - 4}.\]")
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Please share your work/thoughts about this assignment.
Looking at the denominators - do you see that x CANNOT be equal to "certain" values?
D
Deleted member 4993
Guest
Please share your work - how did you get those values?
If I were to do this problem- I'll multiply both sides by (x-4) * (x-5)...........(with the restriction that x \(\displaystyle \ne 4 \ \ or \ne 5 \)), to get:
x * (x-4) = 4 * (x-5)
and continue...
[MATH]x=\frac{4x-20}{x-4}[/MATH][MATH](x-4)x=4x-20[/MATH][MATH]x^2-4x=4x-20[/MATH][MATH]x^2-4x-4x+20=0[/MATH][MATH]x^2-4x-4x+20-4=-4[/MATH][MATH](x-4)^2= -4[/MATH][MATH](x-4)(x-4)=-4[/MATH](2)(-2)=-4
(-2)(2)=-4
(-1)(4)=-4
(-4)(1) =-4
so x=
0,2,3,6.
(-2)(2)=-4
(-1)(4)=-4
(-4)(1) =-4
so x=
0,2,3,6.
D
Deleted member 4993
Guest
It should not.
x2 - 8*x + 20 = 0
This is a quadratic equation. Have you been taught about quadratic equation?
topsquark
Senior Member
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Using the quadratic equation I get [math]x = 4 \pm 2i[/math].
You can complete the square, as you were doing:
[math]x^2 - 8x + 20 = 0[/math]
[math]x^2 - 8x + 20 - 4 = -4[/math]
[math](x - 4)^2 = -4[/math]
[math]x - 4 = \sqrt{-4} = \pm 2i[/math]
[math]x = 4 \pm 2i[/math]
I have no idea what you are doing here.
-Dan
Steven G
Elite Member
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I agree with (2)(-2)=-4, (-2)(2)=-4, (-1)(4)=-4, (-4)(1) =-4, although I do not know why you stating this. I am extremely curious how did you conclude that x must therefore equal to 0, 2, 3 and 6.
firemath
Full Member
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Were you trying to find all the factor pairs here?