I am trying to decipher the steps needed to determine the intersection point between curves represented by the two following equations:
$$y= 2\;x+2$$
$$y= \dfrac{128}{(x+1)^2}$$
I know that they intersect at the point (3,8), but I am unsure of whether I should get rid of the denominator first when starting the process. Could someone please work out the project so I can understand how to arrive at this point?
$\endgroup$ 13 Answers
$\begingroup$One cannot "remove" the denominator of an expression. You want to equate the two expressions and then solve for $x$. So, start with $2x + 2 = \frac{128}{(x+1)^2}$.
$\endgroup$ 4 $\begingroup$Equating the expressions,
$2x+2={128 \over (x+1)^2}\\or, 2(x+1)={128 \over (x+1)^2}\\or, 2(x+1)^3=128\\or,(x+1)^3=64\\or, (x+1)^3-64=0\\or, (x+1)^3-4^3=0\\or,(x+1-4)((x+1)^2+4(x+1)+16)=0\\or,(x-3)(x^2+6x+21)=0\\\text {Either} \space (x-3)=0 \space\text {or} \space x^2+6x+21=0.\\x-3=0 \space \text {gives} \space x=3 $
The quadratic part gives complex roots which are ignored if you are looking for only real solutions.
Then using $y=2x+2 \text {we get} y=2\times3+2=8$
So the point of intersection is $(3,8)$.
$\endgroup$ $\begingroup$Note that $(x+1)$ is common factor . So, set it equal to u.
$$ y= 2 u = 128/u^2 \rightarrow( 2 u^3 = 128) \rightarrow ( u= 4 ) \rightarrow (x=3, y= 8) $$
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