How do you solve rational equations $81.9 = \frac{{(0.1 + 2 x)}^{2}}{(0.1 - x) (0.1 - x)}$ $81.9=((0.1+2x)^2)/((0.1-x)(0.1-x))$ ?

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Answer 1 · 2018-02-08T20:11:15

$x = 0.0488$ $x=0.0488$ or $x = 0.213$ $x=0.213$ (3sf)

$81.9 = \frac{{(0.1 + 2 x)}^{2}}{(0.1 - x) (0.1 - x)}$ $81.9=((0.1+2x)^2)/((0.1-x)(0.1-x)$

We can rewrite the dominator, to get

$81.9 = \frac{{(0.1 + 2 x)}^{2}}{{(0.1 - x)}^{2}}$ $81.9=((0.1+2x)^2)/((0.1-x)^2$

Multiply by ${(0.1 - x)}^{2}$ $(0.1-x)^2$

$81.9 {(0.1 - x)}^{2} = {(0.1 + 2 x)}^{2}$ $81.9(0.1-x)^2=(0.1+2x)^2$

Expand the brackets

$81.9 (0.01 - 0.2 x + x^{2}) = (0.01 + 0.4 x + 4 x^{2})$ $81.9(0.01-0.2x+x^2)=(0.01+0.4x+4x^2)$

$0.819 - 16.38 x + 81.9 x^{2} = 0.01 + 0.4 x + 4 x^{2}$ $0.819-16.38x+81.9x^2=0.01+0.4x+4x^2$

$77.9 x^{2} - 20.38 x + 0.809 = 0$ $77.9x^2-20.38x+0.809=0$

$x = \frac{20.38 \pm \sqrt{{20.38}^{2} - 4 \times 77.9 \times 0.809}}{2 \times 77.9}$ $x=(20.38+-sqrt(20.38^2-4xx77.9xx0.809))/(2xx77.9)$

$x = 0.0488$ $x=0.0488$ or $x = 0.213$ $x=0.213$ (3sf)

How do you solve rational equations 81.9=((0.1+2x)^2)/((0.1-x)(0.1-x))81.9=(0.1+2x)2(0.1−x)(0.1−x)81.9=((0.1+2x)^2)/((0.1-x)(0.1-x))?