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Let equation \(x^2+ax+b=0\) and \(x^2+cx+d=0 \)    (a≠c) have the same root \(x\). Please determine this root value using a, b, c, d to represent it. 

 Sep 24, 2018

we can add the two equations to obtain


\(2x^2 +(a+c)x+(b+d)=0\\ x^2 + \dfrac{a+c}{2}x+\dfrac{b+d}{2}=0\)


\(\text{Now just apply the quadratic formula}\\ r_{1,2} = \dfrac{-\frac{a+c}{2}\pm \sqrt{\left(\frac{a+c}{2}\right)^2 - 2(b+d)}}{2}\)


\(\text{we can clean this up a bit}\\ r_{1,2} = \frac{1}{4} \left(\pm\sqrt{(a+c)^2-8 (b+d)}-a-c\right)\)


\(\text{If there is only a single shared root }r_1 = r_2\\ \text{if there are no shared roots}\\ (a+c)^2 - 8(b+d)<0\)

 Sep 25, 2018

Why add the two equations ?

Why not simply subtract one from the other ?


Suppose that the first equation has roots p and q and that the second equation has roots p and s,

so that the common root is p.

Then we have

\(\displaystyle (x-p)(x-q)=0 \text{ and }(x-p)(x-s)=0\; .\)

Subtracting the second equation from the first,

\(\displaystyle (x-p)\{(x-q)-(x-s)\}=0\; ,\)


\(\displaystyle (x-p)(s-q)=0\; ,\)

showing that  x = p  is the  root of the resulting equation,  \(\displaystyle (\;s \neq q\;)\;.\)


So, going back to the original equations and subtracting one from the other,

\(\displaystyle x(a-c)+(b-d)=0\;,\)

\(\displaystyle x=\frac{d-b}{a-c}\; .\)



 Sep 25, 2018

you're right though I'd do it as


\(x^2+a x+b - (x^2+c x+d) = 0\\ (a-c)x + (b-d) = 0 \\ \forall a \neq c \\ x = \dfrac{d-b}{a-c}\)

Rom  Sep 27, 2018

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