Solve for x: \\ \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}

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Apr 11, 2026, 2:12:39 AMApr 11

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Mathematics for School & Entrance Level • 1.5K followers

This space shows mathematics, suitable for school & Engineering Entrance Exam.

Enrico Gregorio, Associate professor in Algebra

Set [math]s=a+b[/math] and [math]p=ab.[/math] Your equation becomes

[math]\dfrac{1}{x+s}=\dfrac{s}{p}+\dfrac{1}{x}[/math]

hence

[math](x+s)(sx+p)=px[/math]

Expanding and reordering

[math]sx^2+s^2x+sp=0[/math]

Assuming [math]s\ne0[/math] the equation becomes...

Pradeep Hebbar, Many years of Structural Engineering & Math enthusiasm

Given equation

[math]x^2-8x +2=0[/math]

Its roots are [math]m[/math] and [math]n[/math]

[math]m+n= \frac{-(-8)}{1} = 8[/math]

[math]mn= \frac{2}{1} = 2[/math]

Now,

[math]\dfrac{1}{m^2}+\dfrac{1}{n^2} = \dfrac{m^2+n^2}{(mn)^2} = \dfrac{(m+n)^2-2mn}{(mn)^2} = \dfrac{8^2-2(2)}{(2)^2} = 15[/math]

Shambhu Bhat, Retired professor in engineering ;Very fond of mathematics

[math]\text { Let $ a-3,a-2,a-1,a $ be the numbers.}[/math]

[math]10(a-3) +a-2=(a-1)\cdot a \implies [/math]

[math]a^2-12a+32=0 [/math]

[math]\text { The roots are $ 4,8 $}[/math]

.[math]\text { the two sets are ,}[/math]

[math]1,2,3,4\tag *{}[/math]

[math]\boxed{\boldsymbol{5,6,7,8}} \tag*{}[/math]

[math]\text { No more because quadratic equation has only 2 solutions.}[/math]

....

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