Question

solve the equation 1/x+1+2/x+2= 4/x+4 by quadratic formula
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Answer 1

$\bf{\Huge{\boxed{\tt{\green{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

$\sf{\Large{\frac{1}{x+1} \:+\:\frac{2}{x+2} \:=\:\frac{4}{x+4}} }$

$\bf{\Large{\underline{\tt{\pink{Explanation\::}}}}}$

$\longmapsto\sf\frac{1}{x+1} \:+\:\frac{2}{x+2} \:=\:\frac{4}{x+4} }$

$\longmapsto\sf{\frac{1(x+2)+2(x+1)}{(x+1)(x+2)} \:=\:\frac{4}{x+4} }$

$\longmapsto\sf{\frac{x+2+2x+2}{(x+1)(x+2)} \:=\:\frac{4}{x+4} }$

$\longmapsto\sf{\frac{3x+4}{(x+1)(x+2)}\:=\:\frac{4}{x+4}}$

$\longmapsto\sf{\frac{3x+4}{x^{2} +2x+x+2} \:=\:\frac{4}{x+4} }$

$\longmapsto\sf{\frac{3x+4}{x^{2}+3x+2 } \:=\:\frac{4}{x+4} }$

$\longmapsto\sf{(x+4)(3x+4)\:=\:4(x^{2} +3x+2)}$

$\longmapsto\sf{3x^{2} \:+\:4x\:+\:12x\:+\:16\:=\:4x^{2} +12x+8}$

$\longmapsto\sf{3x^{2} +16x+16=4x^{2} +12x+8}$

$\longmapsto\sf{3x^{2} -4x^{2} +16x-12x+16-8\:=\:0}$

$\longmapsto\sf{-x^{2} +4x+8\:=\:0}$

$\longmapsto\sf{-(x^{2} -4x-8)\:=\:0}$

$\longmapsto\sf{\red{x^{2} -4x-8\:=\:0}}$

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Using Sridharacharya formula:

$\leadsto\sf{\Large{\boxed{\pink{x\:=\:\frac{-b \pm \sqrt{b^{2}-4ac } }{2a} }}}}$

Compare this equation with ax²+bx+c=0, we get

• a = 1
• b = -4
• c = -8

Therefore,

$\longmapsto\sf{x\:=\:\frac{-(-4) ± \sqrt{(-4)^{2}-4*1*(-8) } }{2*1} }$

$\longmapsto\sf{x\:=\:\frac{4 ± \sqrt{16+32} }{2} }$

$\longmapsto\sf{x\:=\:\frac{4 ± \sqrt{48} }{2} }$

$\longmapsto\sf{x\:=\:\frac{4 ± 4\sqrt{3} }{2} }$

$\longmapsto\sf{\red{x\:=\:2 ± 2\sqrt{3} }}$

Or

$\mapsto\sf{x\:=\:2+\sqrt{3} \:\:\:\:\:\:or\:\:\:\:\:\:\:x\:=\:2-\sqrt{3} }$