How do you solve $\frac { 4} { x - 3} = \frac { x + 5} { 5} ]$ ?

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Answer 1 · 2018-01-11T17:57:57

$x_1=5$ and $x_2=-7$

$4/(x-3)=(x+5)/5$

Perform this
$4/(x-3)$ χ $(x+5)/5$

$<=>$ $4*5=(x-3)(x+5)$ $<=>$

$20=x^2+5x-3x-15 <=>$

$x^2+2x-35=0$

Discriminant will be $Δ=b^2-4ac=4-4*1*(-35)=140+4=144$

$x_(1,2)=(-b+-sqrt(Δ))/(2α)$ so

$x_1=(-2+12)/2=5$

$x_2=(-2-12)/2=-14/2=-7$

As a result the roots are $x_1=5$ and $x_2=-7$

How do you solve \frac { 4} { x - 3} = \frac { x + 5} { 5} ]\frac { 4} { x - 3} = \frac { x + 5} { 5} ]?