Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$ First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^2-6x+15)\\x^4-12x^3+15x^2-10x^3+120x-150x+15x^2-180x+225=\\=4x^3-24x^2+60x$$ which is an equation I can't solve. I tried to simplify the LHS by $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{(x^2-6x+15)-4x}{x^2-6x+15}=1-\dfrac{4x}{x^2-6x+15}$$ but this isn't helpful at all. Any help would be appreciated! :) Thank you in advance!
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3 Answers
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If you take out $x$ (clearly $x\ne 0$) we get: $$\dfrac{x(x-10+{15\over x})}{x(x-6+{15\over x})}=\dfrac{4x}{x(x-12+{15\over x})}.$$
Cancel $x$ and let $t=x+ {15\over x}$, then we have $${t-10\over t-6} = {4\over t-12}$$ or $$t^2-22t+120 = 4t-24$$ and so on...
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let $p=x^2+15,q=x$ then we have to solve $$\frac{p-10q}{p-6q}=\frac{4q}{p-12q}$$ $$\iff (p-10q)(p-12q)-4q(p-6q)=0$$ $$\iff (p - 8 q) (p - 18 q)=0$$ provided that $p\neq 6q,p\neq 12q$
Can you end it now?
answered Apr 28, 2021 at 15:53
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$\begingroup$ Thank you for the response! May I ask you how did you get $$(p-8q)(p-18q)=0?$$ Did you expand the LHS? $\endgroup$– kormoranApr 28, 2021 at 15:56
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$\begingroup$ @Medi divide the previous equation by $q^2$ and sub $t=\frac{p}{q}$ to get a quadratic and factor! $\endgroup$ Apr 28, 2021 at 15:58
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$\begingroup$ After expanding it /the previous equation/, right? $\endgroup$– kormoranApr 28, 2021 at 15:59
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$\begingroup$ So now we will have $p=8q\Rightarrow x^2+15=8x$ or $p=18q\Rightarrow x^2+15=18x$? $\endgroup$– kormoranApr 28, 2021 at 16:01
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Componendo and dividendo yields:
$$\frac{x^2-6x+15}{x^2-10x+15} = \frac{x^2-12x+15}{4x}$$ $$\Rightarrow \frac{4x}{x^2 - 10x + 15} = \frac{x^2-16x+15}{4x} \tag{$\frac{a-b}{b} = \frac{c-d}{d}$}$$ $$\Rightarrow u=x^2-13x+15: \frac{4x}{u + 3x} = \frac{u - 3x}{4x}$$ $$\Rightarrow u^2 - 9x^2 = 16x^2$$ $$\Rightarrow (u - 5x)(u + 5x) = 0$$ $$\Rightarrow (x^2 - 18x + 15)(x^2 - 8x + 15) = 0$$
