Question

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separated by commas.

Answer 1

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then the equation has no solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}\\\\\text{if}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16<0\\\\\bold{NO\ REAL\ SOLUTIONS}$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$

Answer 2

$\frac{c}{c - 5} = \frac{4}{c - 4}$

$(c - 5)(c - 4) \times \frac{c}{c - 5} = \frac{4}{c - 4} \times (c - 5)(c - 4)$

$(c - 4) \times c = 4(c - 5)$

With c≠4, 5

${c}^{2} - 4c = 4c - 20$

${c}^{2} - 8c + 20 = 0$

$c = \frac{2 - + \sqrt{2 - 20} }{2}$

But 2-20 is negative, therefore there isn't any real solution to this equation.