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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.

User Toffler
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2 Answers

4 votes


(c)/(c - 5) = (4)/(c - 4)


(c - 5)(c - 4) * (c)/(c - 5) = (4)/(c - 4) * (c - 5)(c - 4)


(c - 4) * c = 4(c - 5)

With c≠4, 5


{c}^(2) - 4c = 4c - 20


{c}^(2) - 8c + 20 = 0


c = (2 - + √(2 - 20) )/(2)

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

User Couz
by
8.5k points
3 votes

Answer:


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

Explanation:


Domain:\\c-4\\eq0\ \wedge\ c-5\\eq0\Rightarrow c\\eq4\ \wedge\ c\\eq5\\\\(c)/(c-5)=(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=(-b)/(2a)\\\\\text{if}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=(-b\pm√(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=√(-1)\\\\\text{therefore}\ √(b^2-4ac)=√(-16)=√((16)(-1))=√(16)\cdot√(-1)=4i\\\\x=(-(-8)\pm4i)/(2(1))=(8\pm4i)/(2)=(8)/(2)\pm(4i)/(2)=4\pm2i

User EEEEH
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