+279 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.
+6251 we can add the two equations to obtain
\(2x^2 +(a+c)x+(b+d)=0\\ x^2 + \dfrac{a+c}{2}x+\dfrac{b+d}{2}=0\)
\(\text{Now just apply the quadratic formula}\\ r_{1,2} = \dfrac{-\frac{a+c}{2}\pm \sqrt{\left(\frac{a+c}{2}\right)^2 - 2(b+d)}}{2}\)
\(\text{we can clean this up a bit}\\ r_{1,2} = \frac{1}{4} \left(\pm\sqrt{(a+c)^2-8 (b+d)}-a-c\right)\)
\(\text{If there is only a single shared root }r_1 = r_2\\ \text{if there are no shared roots}\\ (a+c)^2 - 8(b+d)<0\)
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,
\(\displaystyle (x-p)\{(x-q)-(x-s)\}=0\; ,\)
so
\(\displaystyle (x-p)(s-q)=0\; ,\)
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,
\(\displaystyle x(a-c)+(b-d)=0\;,\)
\(\displaystyle x=\frac{d-b}{a-c}\; .\)
Tiggsy
